Serlo: EN: Exercises: Derivatives 4

Criterion for constancy

Exercise (Simple application)

Let $K>0$ and $\alpha >1$ . For the function $f:D\to \mathbb {R}$ we assume

|f(x)-f(y)|\leq K|x-y|^{\alpha }

for all $x,y\in D$ . Show that then, $f$ is constant.

Solution (Simple application)

By assumption there is

0\leq \left|{\frac {f(x)-f(y)}{x-y}}\right|={\frac {|f(x)-f(y)|}{|x-y|}}\leq {\frac {K|x-y|^{\alpha }}{|x-y|}}=K|x-y|^{\alpha -1}

Further there is $\lim _{x\to y}K|x-y|^{\alpha -1}=0$ , since $\alpha -1>0$ . The squeeze theorem then yields

\lim _{x\to y}\left|{\frac {f(x)-f(y)}{x-y}}\right|=0

for all $y\in D$ . By the computation rule for limits we then get a zero differential quotient

\lim _{x\to y}{\frac {f(x)-f(y)}{x-y}}=0

for all $y\in D$ . Hence $f$ is constant.

Exercise (Proof of identities)

Show that

$\cosh ^{2}(x)-\sinh ^{2}(x)=1$
$\arccos(x)+\arcsin(x)={\frac {\pi }{2}}$ for all $x\in (-1,1)$

Proof (Proof of identities)

Part 1: The function

f:\mathbb {R} \to \mathbb {R} ,\ f(x)=\cosh ^{2}(x)-\sinh ^{2}(x)

is differentiable by the chain- and difference rule with

f'(x)=2\cosh(x)\sinh(x)-2\sinh(x)\cosh(x)=0

So $f\equiv c$ is constant. Since further there is

f(0)=\cosh(0)-\sinh(0)=1-0=1

we have $f(x)=\cosh ^{2}(x)-\sinh ^{2}(x)=c=1$ .

Part 2:

g:(-1,1)\to \mathbb {R} ,\ g(x)=\arcsin(x)+\arccos(x)

is differentiable according to the sum rule, since the arcus-functions are differentiable. Further there is

g'(x)={\frac {1}{\sqrt {1-x^{2}}}}-{\frac {1}{\sqrt {1-x^{2}}}}=0

Hence $g\equiv c$ is constant. Further there is

f(0)=\arcsin(0)+\arccos(0)=0+{\frac {\pi }{2}}={\frac {\pi }{2}}

since $\sin(0)=0$ and $\cos({\tfrac {\pi }{2}})=0$ . So $c={\tfrac {\pi }{2}}$ establishing the assertion.

Exercise (Logarithm representations of ${\text{arcosh}}$ and ${\text{artanh}}$ )

Show that

${\text{arcosh}}(x)=\ln \left(x+{\sqrt {x^{2}-1}}\right)$ for $x>1$
${\text{artanh}}(x)={\frac {1}{2}}\ln \left({\frac {x+1}{x-1}}\right)$ for $|x|<1$

Proof (Logarithm representations of ${\text{arcosh}}$ and ${\text{artanh}}$ )

Part 1:

The function $f={\text{arcosh}}:(1,\infty )\to \mathbb {R}$ is differentiable, see examples for derivatives, with

{\text{arcosh}}'(x)={\frac {1}{\sqrt {x^{2}-1}}}

By the chain- and sum rule also $g:(1,\infty )\to \mathbb {R} ,\ g(x)=\ln \left(x+{\sqrt {x^{2}-1}}\right)$ is differentiable with

g'(x)={\frac {1}{x+{\sqrt {x^{2}-1}}}}\cdot \left(1+{\frac {2x}{2{\sqrt {x^{2}-1}}}}\right)={\frac {1}{x+{\sqrt {x^{2}-1}}}}\cdot \left({\frac {{\sqrt {x^{2}-1}}+x}{\sqrt {x^{2}-1}}}\right)={\frac {1}{\sqrt {x^{2}-1}}}

So we get $f(x)=g(x)+c$ . But now,

f(1)={\text{arcosh}}(1)=0

since $\cosh(0)=1$ , and there is

g(1)=\ln(1)=0

So $c=0$ , and hence $f=g$ .

Part 2:

$f={\text{artanh}}:(-1,1)\to \mathbb {R}$ is differentiable, as well, with

{\text{artanh}}'(x)={\frac {1}{1-x^{2}}}

By the factor-, chain- and quotient rule, also $g:(-1,1)\to \mathbb {R} ,\ g(x)={\frac {1}{2}}\ln \left({\frac {x+1}{x-1}}\right)$ is differentiable with

g'(x)={\frac {1}{2}}{\frac {1}{\frac {x+1}{x-1}}}\cdot {\frac {1\cdot (x-1)-(x+1)\cdot 1}{(x-1)^{2}}}={\frac {1}{2}}{\frac {x-1}{x+1}}\cdot {\frac {-2}{(x-1)^{2}}}={\frac {-1}{x^{2}-1}}={\frac {1}{1-x^{2}}}

So we have $f(x)=g(x)+c$ . Since

f(0)={\text{artanh}}(0)=0

by $\tanh(0)=0$ , as well as

g(0)=\ln(1)=0

there is again $c=0$ , and hence $f=g$ .

Exercise (Extension of the identity theorem)

Let $f,g:[a,b]\to \mathbb {R}$ be twice differentiable with $f''=g''$ . Then $f$ and $g$ differ only by a linear function $x\mapsto cx+d$ with $c,d\in \mathbb {R}$ .

Solution (Extension of the identity theorem)

Because $f''=g''$ according to the identity theorem there is a $c\in \mathbb {R}$ with

f'(x)=\underbrace {g'(x)+c} _{=h'(x)}

if we now set $h:[a,b]\to \mathbb {R} ,\ h(x)=g(x)+cx$ , then there is

h'(x)=g'(x)+c

the identity theorem again provides us with a $d\in \mathbb {R}$ such that

f(x)=h(x)+d=g(x)+cx+d

Exercise (General solution of a differential equation)

Let $f,g:\mathbb {R} \to \mathbb {R}$ be differentiable and $\omega \in \mathbb {R}$ . Further let

{\begin{aligned}f'&=\omega g\\g'&=-\omega f\end{aligned}}

Show that:

$f(x)=a\sin(\omega x)-b\cos(\omega x)$ and $g(x)=a\cos(\omega x)+b\sin(x)$ satisfy the differential equations.
If two functions $f$ and $g$ satisfy the differential equations, then there is $f(x)=a\sin(\omega x)-b\cos(\omega x)$ and $g(x)=a\cos(\omega x)+b\sin(x)$ .
Furthermore, if $\omega =1$ and there is $f(0)=0$ and $g(0)=1$ , then $f=\sin$ and $g=\cos$ .

Proof (General solution of a differential equation)

Part 1: There is

f'(x)=a\omega \cos(\omega x)-b\omega (-\sin(\omega x))=\omega (a\cos(\omega x)+b\sin(\omega x))=\omega g(x)

and

g'(x)=a\omega (-\sin(\omega x))+b\omega \cos(\omega x)=-\omega (a\sin(\omega x)-b\cos(\omega x))=-\omega f(x)

So $f$ and $g$ satisfy the differential equations.

Part 2:We define (as given in the hint) the auxiliary functions

{\begin{aligned}h_{1}:\mathbb {R} \to \mathbb {R} ,\ h_{1}(x)&=f(x)\sin(\omega x)+g(x)\cos(\omega x)\\h_{2}:\mathbb {R} \to \mathbb {R} ,\ h_{2}(x)&=f(x)\cos(\omega x)-g(x)\sin(\omega x)\end{aligned}}

These are differentiable by the product-, sum- and difference rule with

{\begin{aligned}h_{1}'(x)&=\underbrace {f'(x)} _{=\omega g(x)}\sin(\omega x)+f(x)\omega \cos(\omega x)+\underbrace {g'(x)} _{=-\omega f(x)}\cos(\omega x)-g(x)\omega \sin(\omega x)\\&=\omega g(x)\sin(\omega x)+\omega f(x)\cos(\omega x)-\omega f(x)\cos(\omega x)-\omega g(x)\sin(\omega x)\\&=0\end{aligned}}

and

{\begin{aligned}h_{2}'(x)&=\underbrace {f'(x)} _{=\omega g(x)}\cos(\omega x)-f(x)\omega \sin(\omega x)-\underbrace {g'(x)} _{=-\omega f(x)}\sin(\omega x)-g(x)\omega \cos(\omega x)\\&=\omega g(x)\cos(\omega x)-\omega f(x)\sin(\omega x)+\omega f(x)\sin(\omega x)-\omega g(x)\cos(\omega x)\\&=0\end{aligned}}

by the criterion for constancy, there is now

{\begin{aligned}h_{1}(x)&=f(x)\sin(\omega x)+g(x)\cos(\omega x)=a\\h_{2}(x)&=f(x)\cos(\omega x)-g(x)\sin(\omega x)=b\end{aligned}}

with $a,b\in \mathbb {R}$ . Further there is

{\begin{aligned}\sin(\omega x)h_{1}(x)+\cos(\omega x)h_{2}(x)&=\sin(\omega x)[f(x)\sin(\omega x)+g(x)\cos(\omega x)]+\cos(\omega x)[f(x)\cos(\omega x)-g(x)\sin(\omega x)]=a\sin(\omega x)+b\cos(\omega x)\\\iff &f(x)\sin ^{2}(\omega x)+g(x)\sin(\omega x)\cos(\omega x)+f(x)\cos ^{2}(\omega x)-g(x)\sin(\omega x)\cos(\omega x)=a\sin(\omega x)+b\cos(\omega x)\\\iff &f(x)\underbrace {(\sin ^{2}(\omega x)+\cos ^{2}(\omega x))} _{=1}=a\sin(\omega x)+b\cos(\omega x)\\&\\\cos(\omega x)h_{1}(x)-\sin(\omega x)h_{2}(x)&=\cos(\omega x)[f(x)\sin(\omega x)+g(x)\cos(\omega x)]-\sin(\omega x)[f(x)\cos(\omega x)-g(x)\sin(\omega x)]=a\cos(\omega x)-b\sin(\omega x)\\\iff &f(x)\sin(\omega x)\cos(\omega x)+g(x)\cos ^{2}(\omega x)-f(x)\sin(\omega x)\cos(\omega x)+g(x)\sin ^{2}(\omega x)=a\cos(\omega x)-b\sin(\omega x)\\\iff &g(x)\underbrace {(\cos ^{2}(\omega x)+\sin ^{2}(\omega x))} _{=1}=a\cos(\omega x)-b\sin(\omega x)\end{aligned}}

So $f(x)=a\sin(\omega x)+b\cos(\omega x)$ and $g(x)=a\cos(\omega x)-b\sin(\omega x)$ .

Part 3: If further $\omega =1$ and

{\begin{aligned}f(0)&=a\sin(0)+b\cos(0)=a\cdot 0+b\cdot 1=b=0\\g(0)&=a\cos(0)-b\sin(0)=a\cdot 1-b\cdot 0=a=1\end{aligned}}

then

{\begin{aligned}f(x)&=1\cdot \sin(x)+0\cdot \cos(x)=\sin(x)\\g(x)&=1\cdot \cos(x)-0\cdot \sin(x)=\cos(x)\end{aligned}}

Hint

Since $f''=(f')'=(\omega g)'=\omega g'=-\omega ^{2}f$ and analogously $g''=-\omega ^{2}g$ both the functions $f$ and $g$ satisfy the differential equation $f''+\omega f=0$ (or $g''+\omega g=0$ ).

Monotony criterion

Exercise (Monotony of the exponential function)

Show using the monotony criterion that:

For all $x>0$ there is $\ln(1+{\tfrac {1}{x}})-{\tfrac {1}{x+1}}>0$
$f:\mathbb {R} ^{+}\to \mathbb {R} ,\ f(x)=\left(1+{\tfrac {1}{x}}\right)^{x}$ is strictly monotonously increasing.

Hint: use 1. in order to prove 2.

Solution (Monotony of the exponential function)

Part 1: For the differentiable auxiliary function $h:\mathbb {R} ^{+}\to \mathbb {R} ,\ h(x)=\ln \left(1+{\tfrac {1}{x}}\right)-{\frac {1}{x+1}}$ there is

h'(x)={\frac {1}{1+{\tfrac {1}{x}}}}\cdot (-{\tfrac {1}{x^{2}}})-{\frac {-1}{(x+1)^{2}}}=-{\frac {\frac {1}{x^{2}}}{\frac {x+1}{x}}}+{\frac {1}{(x+1)^{2}}}=-{\frac {1}{x(x+1)}}+{\frac {1}{(x+1)^{2}}}={\frac {-(x+1)-x}{x(x+1)^{2}}}=-{\frac {1}{x(x+1)^{2}}}<0

So $h$ is strictly monotonously decreasing by the monotony criterion. Further

Graph of the function

x\mapsto \ln(1+{\tfrac {1}{x}})-{\tfrac {1}{x+1}}

\lim _{x\to 0+}h(x)=\lim _{x\to 0+}\underbrace {\ln \left(1+{\tfrac {1}{x}}\right)} _{\to \infty }-\underbrace {\frac {1}{x+1}} _{\to 0}=\infty

and

\lim _{x\to \infty }h(x)=\lim _{x\to \infty }\underbrace {\ln \left(1+{\tfrac {1}{x}}\right)} _{\to \ln(1)=0}-\underbrace {\frac {1}{x+1}} _{\to 0}=0

Since $h$ is continuous and strictly monotonously decreasing, there must be $h(x)=\ln(1+{\tfrac {1}{x}})-{\tfrac {1}{x+1}}>0$ .

Part 2:

There is $f(x)=\left(1+{\tfrac {1}{x}}\right)^{x}=\exp \left(x\ln(1+{\tfrac {1}{x}})\right)$ . Since $\exp$ is strictly monotonously increasing, the function $f$ is strictly monotonously increasing, if and only if the "inner function" $g(x)=x\ln(1+{\tfrac {1}{x}})$ is. This function in turn is differentiable on all of $\mathbb {R} ^{+}$ by the product rule and there is

g'(x)=1\cdot \ln(1+{\tfrac {1}{x}})+x\cdot {\frac {1}{1+{\frac {1}{x}}}}\cdot (-{\tfrac {1}{x^{2}}})=\ln(1+{\tfrac {1}{x}})-{\frac {\frac {1}{x}}{\frac {x+1}{x}}}=\underbrace {\ln(1+{\tfrac {1}{x}})-{\frac {1}{x+1}}} _{=h(x){\text{ from 1.}}}>0

By the monotony criterion, the function $g$ , and hence also $f$ is strictly monotonously increasing.

Exercise (Condition for monotonicity of a cubic function)

Let $a,b,c,d\in \mathbb {R}$ . Provide a condition for $a,b,c,d$ such that

f:\mathbb {R} \to \mathbb {R} ,\ f(x)=ax^{3}+bx^{2}+cx+d

is strictly monotonously increasing on all of $\mathbb {R}$ .

Hint: Distinguish the cases $a=0$ , $a>0$ and $a<0$

Solution (Condition for monotonicity of a cubic function)

For $f$ being strictly monotonously increasing on all of $\mathbb {R}$ , we need by the monotony criterion that

f'(x)=3ax^{2}+2bx+c>0

holds for all $x\in \mathbb {R}$ .

Fall 1: $a=0$

Then $f'(x)=2bx+c$ . For $f$ to be strictly monotonously increasing, $f'(x)=2bx+c>0\iff 2bx>-c$ must hold. For $b\neq 0$ this is never possible for any $x\in \mathbb {R}$ .

However, if $b=0$ , then there is $f'(x)>0\iff c>0$ . So $f$ is strictly monotonously increasing for $a=b=0$ and $c>0$ .

Fall 2: $a>0$

By completing the square, we get

{\begin{aligned}f'(x)&=3ax^{2}+2bx+c=3a(x^{2}+{\tfrac {2b}{a}}x)+c\\[0.3em]&=3a(x^{2}+{\tfrac {2b}{3a}}x+({\tfrac {b}{3a}})^{2}-({\tfrac {b}{3a}})^{2})+c\\[0.3em]&=3a(x+{\tfrac {b}{3a}})^{2}-{\tfrac {b^{2}}{3a}}+c\\[0.3em]&=3a(x+{\tfrac {b}{3a}})^{2}-{\tfrac {b^{2}-3ac}{3a}}\end{aligned}}

So $f$ is strictly monotonously increasing whenever there is

{\begin{aligned}&3a(x+{\tfrac {b}{3a}})^{2}-{\tfrac {b^{2}-3ac}{3a}}>0\\\iff &3a(x+{\tfrac {b}{3a}})^{2}>{\tfrac {b^{2}-3ac}{3a}}\\{\overset {3a>0}{\iff }}&(x+{\tfrac {b}{3a}})^{2}>{\tfrac {b^{2}-3ac}{(3a)^{2}}}\\\end{aligned}}

This is satisfied for all $x\in \mathbb {R}$ if and only if the right-hand side ${\tfrac {b^{2}-3ac}{(3a)^{2}}}$ is negative. This in turn is exactly the case for

b^{2}-3ac<0\iff b^{2}<3ac

Hence, $f$ is strictly monotonously increasing for $a>0$ and $3ac>b^{2}$ .

Fall 3: $x>0$

Here, we have

{\begin{aligned}f'(x)>0\iff &3a(x+{\tfrac {b}{3a}})^{2}>{\tfrac {b^{2}-3ac}{3a}}\\{\overset {3a<0}{\iff }}&(x+{\tfrac {b}{3a}})^{2}<{\tfrac {b^{2}-3ac}{(3a)^{2}}}\end{aligned}}

However, this is never fulfilled for all $x\in \mathbb {R}$ . So in this case $f$ is never strictly monotonous increasing.

Hint

Similarly, we can show that $f$ is strictly monotonously decreasing in the cases $a=b=0$ and $c<0$ , as well as for $a<0$ and $3ac>b^{2}$ .

Exercise (Applying the monotony criterion)

Let $f:[0,1]\to \mathbb {R}$ be differentiable with $f(0)=0$ . Further let $f'(x)\leq \lambda f(x)$ for some (fixed) $\lambda >0$ and all $x\in [0,1]$ . Show that there is

f(x)\leq 0

for all

x\in [0,1]

Hint: Consider the auxiliary function $h(x)=f(x)e^{-\lambda x}$ .

Proof (Applying the monotony criterion)

As stated in the hint we consider

h:[0,1]\to \mathbb {R} ,\ h(x)=f(x)e^{-\lambda x}

$h$ is differentiable according to the product rule with

h'(x)=f'(x)e^{-\lambda x}+f(x)e^{-\lambda x}(-\lambda )=e^{-\lambda x}(f'(x)-\lambda f(x))

But now $e^{-\lambda x}>0$ and by assumption $f'(x)\leq \lambda f(x)\iff f'(x)-\lambda f(x)\leq 0$ . So there is

h'(x)\leq 0

for all

x\in [0,1]

by the monotony criterion, $h$ is monotonously decreasing. Since further there is $h(0)=f(0)e^{0}=f(0)=0$ we have

h(x)\leq h(0)=0

for all

x\in [0,1]

Therefore, we also have

f(x)=\underbrace {h(x)} _{\leq 0}\underbrace {e^{\lambda x}} _{>0}\leq 0

for all

x\in [0,1]

Derivative and extrema

Exercise (Extrema of functions 1)

Investigate whether the following functions have local/global extrema. Determine and characterise these if they exist.

$f:(0,\infty )\to \mathbb {R} ,\ f(x)={\tfrac {(\ln(x))^{2}}{x}}$
$g:[{\tfrac {1}{2}},\infty )\to \mathbb {R} ,\ g(x)={\tfrac {1}{\arctan(2x)}}$

Solution (Extrema of functions 1)

Part 1:

Part 1: Local extrema of $f$

$f$ is differentiable on $(0,\infty )$ according to the quotient rule with

f'(x)={\frac {2\ln(x){\frac {1}{x}}\cdot x-(\ln(x))^{2}\cdot 1}{x^{2}}}={\frac {\ln(x)(2-\ln(x))}{x^{2}}}

According to the sufficient criterion for the existence of an extremum ${\tilde {x}}\in \mathbb {R} ^{+}$ , there must be $f'({\tilde {x}})=0$ . Now

f'(x)={\frac {\ln(x)(2-\ln(x))}{x^{2}}}=0\iff \ln(x)=0{\text{ or }}\ln(x)=2\iff x=1{\text{ or }}x=e^{2}

So ${\tilde {x}}=1$ and ${\hat {x}}=e^{2}$ are the candidates for local extrema in $(0,\infty )$ . Now there is

f(x)={\frac {\ln(x)(2-\ln(x))}{x^{2}}}>0\iff \ln(x)>0{\text{ and }}2-\ln(x)>0\iff x>1{\text{ and }}x<e^{2}

The case $\ln(x)<0\iff x<1$ and $2-\ln(x)<0\iff x>e^{2}$ is not possible. Thus $f'(x)>0$ holds on $(1,e^{2})$ .

Further there is

f(x)={\frac {\ln(x)(2-\ln(x))}{x^{2}}}<0\iff {\begin{cases}\ln(x)>0{\text{ and }}2-\ln(x)<0\iff x>1{\text{ and }}x>e^{2}\iff x>e^{2},&{\text{ or }}\\\ln(x)<0{\text{ and }}2-\ln(x)>0\iff x<1{\text{ and }}x<e^{2}\iff x<1&\end{cases}}

So $f'(x)<0$ holds on $(0,1)$ and on $(e^{2},\infty )$ .

By the sufficient criterion, ${\tilde {x}}=1$ is a (strict) local minimum and ${\hat {x}}=e^{2}$ is a (strict) local maximum of $f$ .

Part 2: Global extrema of $f$

For global extrema we first have to determine the limits $\lim _{x\to 0+}f(x)$ and $\lim _{x\to \infty }f(x)$ .

Since $\lim _{x\to 0+}\ln(x)=-\infty$ and $\lim _{x\to 0+}{\tfrac {1}{x}}=\infty$ there is

Graph of the function

f

\lim _{x\to 0+}f(x)=\lim _{x\to 0+}{\frac {(\ln(x))^{2}}{x}}=\infty

Thus $f$ is unbounded from above, and therefore has no local extremum. Further, for $x\to \infty$ every power of $\ln$ grows slower than $x$ . Thus

\lim _{x\to \infty }f(x)=\lim _{x\to \infty }{\frac {(\ln(x))^{2}}{x}}=0+

(As numerator and denominator are positive.) Now $f({\tilde {x}})=f(1)={\tfrac {(\ln(1))^{2}}{1}}=0$ . Thus ${\tilde {x}}=1$ is a global minimum of $f$ .

Part 2:

Part 1: Local extrema of $g$

$g$ is differentiable on $({\tfrac {1}{2}},\infty )$ by the chain rule with

Graph of the function

g

g'(x)={\frac {-1}{(\arctan(2x))^{2}}}\cdot {\frac {1}{1+(2x)^{2}}}\cdot 2=-{\frac {2}{(\arctan(2x))^{2}(1+4x^{2}))}}

Since now $(\arctan(2x))^{2}>0$ and $1+4x^{2}>0$ there is, $g'(x)<0$ . By the necessary criterion for extrema, $g$ has no local extrema on $({\tfrac {1}{2}},\infty )$ .

Since $g$ is continuous on $[{\tfrac {1}{2}},\infty )$ , it follows from $g'(x)<0$ for all $x\in ({\tfrac {1}{2}},\infty )$ that $g$ is strictly monotonously decreasing on $[{\tfrac {1}{2}},\infty )$ . Therefore $g$ has a local maximum at ${\tilde {x}}={\tfrac {1}{2}}$ .

Part 2: Global extrema of $g$

Using the same argument as in part 1, it follows that ${\tilde {x}}={\tfrac {1}{2}}$ is even a global maximum of $g$ .

Exercise (Extrema of functions 2)

Investigate whether the following functions are continuous, differentiability and/or have local/global extrema:

f:\mathbb {R} \to \mathbb {R} ,\ f(x)={\begin{cases}x^{x}&{\text{ for }}x>0,\\x+1&{\text{ for }}x\leq 0\end{cases}}

Solution (Extrema of functions 2)

Part 1: continuity and differentiability

Continuity:

On $(-\infty ,0)$ the function $f(x)=x+1$ is continuous as a polynomial. The function $f(x)=x^{x}=\exp(x\ln(x))$ is continuous on $(0,\infty )$ as a composition of the continuous functions $\exp$ , ${\text{id}}$ and $\ln$ . At $x=0$ there is

{\begin{aligned}\lim _{x\to 0-}f(x)&=\lim _{x\to 0-}x+1=0+1=1,{\text{ and }}\\[0.3em]f(0)&=0+1=1\end{aligned}}

In addition, since $\lim _{x\to 0+}x\ln(x)=0$ and by continuity of the exponential function

\lim _{x\to 0+}f(x)=\lim _{x\to 0+}x^{x}=\lim _{x\to 0+}\exp(x\ln(x))=\exp(\lim _{x\to 0+}x\ln(x))=\exp(0)=1

So $f$ is continuous at zero and hence on all of $\mathbb {R}$ .

Differentiability:

On $(-\infty ,0)$ the function $f(x)=x+1$ is differentiable as a polynomial, with

f'(x)=1

On $(0,\infty )$ the function $f(x)=x^{x}=\exp(x\ln(x))$ is differentiable by the chain- and product rule as it is a composition of differentiable functions $\exp$ , ${\text{id}}$ and $\ln$ . There is

f'(x)=\exp(x\ln(x))\cdot (1\cdot \ln(x)+x\cdot {\frac {1}{x}})=x^{x}(\ln(x)+1)

At $x=0$ there is, according to L'Hospital's rule,

\lim _{x\to 0+}{\frac {f(x)-f(0)}{x-0}}=\lim _{x\to 0+}{\frac {x^{x}-1}{x}}{\underset {\text{l.H.}}{\overset {\frac {0}{0}}{=}}}\lim _{x\to 0+}{\frac {x^{x}(\ln(x)+1)}{1}}{\overset {\frac {-\infty }{1}}{=}}-\infty

So $f$ is not differentiable at zero.

Part 2: Local and global extrema

Local extrema:

On $(-\infty ,0)$ there is $f'(x)=1\neq 0$ . So $f$ cannot have local extrema there.

On $(0,\infty )$ however

f'(x)=x^{x}(\ln(x)+1)=\underbrace {\exp(x\ln(x))} _{\neq 0}(\ln(x)+1)=0\iff \ln(x)=-1\iff x=e^{-1}={\frac {1}{e}}

So ${\tilde {x}}={\tfrac {1}{e}}$ is a candidate for a possible extremum. Further

f'(x)=\underbrace {x^{x}} _{>0}(\ln(x)+1){\begin{cases}>0\iff \ln(x)+1>0\iff \ln(x)>-1\iff x>{\tfrac {1}{e}},\\<0\iff \ln(x)+1<0\iff \ln(x)<-1\iff x<{\tfrac {1}{e}}\end{cases}}

So $f$ has a strict local minimum in ${\tilde {x}}={\tfrac {1}{e}}$ .

Now we still have to examine ${\hat {x}}=0$ . Since $f$ is not differentiable there, our necessary and sufficient criteria are not applicable. However, there is

f'(x)=1>0

for all

x\in (-\infty ,0)

and

f'(x)=x^{x}(\ln(x)+1)<0

for all

x\in (0,{\tfrac {1}{e}})

Thus $f$ is strictly monotonously increasing on $(-\infty ,0)$ , and strictly monotonously decreasing on $(0,{\tfrac {1}{e}})$ . Since $f$ is continuous at zero, it follows that

f(0)>f(x)

for all

x\in (-{\tfrac {1}{e}},{\tfrac {1}{e}})\setminus \{0\}

So $f$ has a strict local maximum in ${\hat {x}}=0$ .

Global extrema:

There is

Graph of the function

f

\lim _{x\to -\infty }f(x)=\lim _{x\to -\infty }x+1=-\infty

and

\lim _{x\to \infty }f(x)=\lim _{x\to \infty }x^{x}=\lim _{x\to \infty }\exp(\underbrace {x\ln(x)} _{\to \infty })=\infty

Therefore $f$ is unbounded from above and below, and has no global extrema.

Exercise (Extrema of functions 3)

Show that the function

f:\mathbb {R} ^{+}\to \mathbb {R} ,\ f(x)=e^{3x}\ln(x)

has exactly two local extrema, and determine their type.

Solution (Extrema of functions 3)

Candidates for the extreme values are obtained from our necessary condition

f'(x){\underset {\text{rule}}{\overset {\text{Prod.-}}{=}}}3e^{3x}\ln(x)+e^{3x}{\frac {1}{x}}=\underbrace {\frac {3e^{3x}}{x}} _{>0}(\underbrace {x\ln(x)+{\frac {1}{3}}} _{:=h(x)}){\overset {!}{=}}0

Graph of the auxiliary function

h

Since the zeros of $h$ cannot be calculated explicitly, we need to examine this function more closely. There is

$h'(x)=\ln(x)+1{\begin{cases}>0\iff x>{\tfrac {1}{e}},\\<0\iff x<{\tfrac {1}{e}}\end{cases}}$
$\lim \limits _{x\to 0}h(x)={\frac {1}{3}}$ , $\lim \limits _{x\to \infty }h(x)=\infty$ and $h({\tfrac {1}{e}})=\underbrace {-{\tfrac {1}{e}}} _{<-{\tfrac {1}{3}}}+{\tfrac {1}{3}}<0$

Because of continuity and 2. $h$ with the intermediate value theorem has (at least) two zeros $x_{1}\in (0,{\tfrac {1}{e}})$ and $x_{2}\in ({\tfrac {1}{e}},\infty )$ .

Because of 1., the function $h$ is strictly monotonously increasing on $(0,{\tfrac {1}{e}})$ and strictly monotonously decreasing on $({\tfrac {1}{e}},\infty )$ . Thus $h$ is respectively injective on $(0,{\tfrac {1}{e}})$ and $({\tfrac {1}{e}},\infty )$ and thus has exactly the two zeros $x_{1}$ and $x_{2}$ .

For the derivative of $f$ we now have

f'(x)={\frac {3e^{3x}}{x}}\cdot h(x)\ {\begin{cases}>0&{\text{ for }}x\in (0,x_{1})\cup (x_{2},\infty ),\\<0&{\text{ for }}x\in (x_{1},x_{2}).\end{cases}}

According to our first sufficient criterion, $f$ has a strict local maximum at $x_{1}$ and a strict local minimum at $x_{2}$ .

Computing limits via L'Hospital

Exercise (L'Hospital 1)

Compute the following limits

$\lim _{x\to 0}{\frac {\sinh(x)}{x}}$
$\lim _{x\to {\frac {\pi }{2}}}{\frac {\sin(x)}{x^{2}}}$
$\lim _{x\to 0+}{\frac {\sin(x)}{x^{2}}}$
$\lim _{x\to 0}{\frac {\sinh(x)}{x^{2}}}$
$\lim _{x\to \infty }{\frac {\ln(x^{p}+1)}{\ln(x^{q})}}$ with $p,q\in \mathbb {R} ^{+}$
$\lim _{x\to \infty }{\frac {\sinh(x)}{\cosh(x)}}$
$\lim _{x\to \infty }{\frac {1-\cos(x)}{x^{2}}}$
$\lim _{x\to 0}{\frac {x^{3}-x^{2}}{\tan(x)}}$
$\lim _{x\to 0+}{\frac {3^{x}-2^{x}}{\tan(x)}}$
$\lim _{x\to 0}{\frac {e^{x}-1-x}{x(e^{x}-1)}}$

Solution (L'Hospital 1)

Part 1:

\lim _{x\to 0}{\frac {\sinh(x)}{x}}\ {\underset {\text{L.H.}}{\overset {\frac {0}{0}}{=}}}\ \lim _{x\to 0}{\frac {\cosh(x)}{1}}={\frac {\cosh(0)}{1}}={\frac {1}{1}}=1

Part 2:

L'Hospital's rule is not applicable here. However, the function $x\mapsto {\tfrac {\sin(x)}{x}}$ is continuous at the point $x={\tfrac {\pi }{2}}$ , and therefore there is

\lim _{x\to {\frac {\pi }{2}}}{\frac {\sin(x)}{x^{2}}}={\frac {\sin({\tfrac {\pi }{2}})}{\left({\tfrac {\pi }{2}}\right)^{2}}}={\frac {1}{\frac {\pi ^{2}}{4}}}={\frac {4}{\pi ^{2}}}

Part 3:

\lim _{x\to 0+}{\frac {\sin(x)}{x^{2}}}\ {\underset {\text{L.H.}}{\overset {\frac {0}{0}}{=}}}\ \lim _{x\to 0+}{\frac {\overbrace {\cos(x)} ^{\to \cos(0)=1}}{2x}}{\overset {\frac {1}{0+}}{=}}+\infty

Part 4:

This limit value does not exist. First it can be decomposed into

\lim _{x\to 0}{\frac {\sinh(x)}{x^{2}}}=\lim _{x\to 0}{\frac {\sinh(x)}{x}}\cdot {\frac {1}{x}}

For the left-hand limit there is now with Part 1:

\lim _{x\to 0-}\underbrace {\frac {\sinh(x)}{x}} _{\to 1}\cdot \underbrace {\frac {1}{x}} _{\to -\infty }=-\infty

Analogously, however, for the right-hand limit:

\lim _{x\to 0+}\underbrace {\frac {\sinh(x)}{x}} _{\to 1}\cdot \underbrace {\frac {1}{x}} _{\to +\infty }=+\infty

So $\lim \limits _{x\to 0-}{\frac {\sinh(x)}{x^{2}}}\neq \lim \limits _{x\to 0+}{\frac {\sinh(x)}{x^{2}}}$ , and hence $\lim _{x\to 0}{\frac {\sinh(x)}{x^{2}}}$ does not exist.

Part 5:

\lim _{x\to \infty }{\frac {\ln(x^{p}+1)}{\ln(x^{q})}}\ {\underset {\text{L.H.}}{\overset {\frac {\infty }{\infty }}{=}}}\ \lim _{x\to \infty }{\dfrac {\dfrac {px^{p-1}}{x^{p}+1}}{\dfrac {qx^{q-1}}{x^{q}}}}=\lim _{x\to \infty }{\dfrac {px^{q}x^{p-1}}{qx^{q-1}(x^{p}+1)}}=\lim _{x\to \infty }{\dfrac {px^{q-1}x^{p}}{qx^{q-1}(x^{p}+1)}}=\lim _{x\to \infty }{\frac {p}{q}}\cdot {\dfrac {x^{p}}{x^{p}+1}}\ {\underset {|:x^{p}}{\overset {|:x^{p}}{=}}}\ \lim _{x\to \infty }{\frac {p}{q}}\cdot {\dfrac {1}{1+\underbrace {\frac {1}{x^{p}}} _{\to 0}}}={\frac {p}{q}}

Part 6:

L'Hospital can be applied here, but it is useless:

\lim _{x\to \infty }{\frac {\sinh(x)}{\cosh(x)}}\ {\underset {\text{L.H.}}{\overset {\frac {\infty }{\infty }}{=}}}\ \lim _{x\to \infty }{\frac {\cosh(x)}{\sinh(x)}}\ {\underset {\text{L.H.}}{\overset {\frac {\infty }{\infty }}{=}}}\ \lim _{x\to \infty }{\dfrac {\sinh(x)}{\cosh(x)}}=\ldots

Instead, it makes sense to use the definitions of $\sinh$ and $\cosh$ , and then transform the quotient:

\lim _{x\to \infty }{\frac {\sinh(x)}{\cosh(x)}}\ {\overset {\text{Def.}}{=}}\ \lim _{x\to \infty }{\dfrac {\frac {e^{x}-e^{-x}}{2}}{\frac {e^{x}+e^{-x}}{2}}}=\lim _{x\to \infty }{\dfrac {2(e^{x}-e^{-x})}{2(e^{x}+e^{-x})}}=\lim _{x\to \infty }{\dfrac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}\ {\underset {|:e^{x}}{\overset {|:e^{x}}{=}}}\ \lim _{x\to \infty }{\dfrac {1-\overbrace {e^{-2x}} ^{\to 0}}{1+\underbrace {e^{-2x}} _{\to 0}}}={\frac {1-0}{1+0}}=1

Part 7:

L'Hospital cannot be applied here because the enumerator $1-\cos(x)$ for $x\to \infty$ diverges (improperly). Instead, the fraction can be estimated as follows:

0\leq \left|{\frac {1-\cos(x)}{x^{2}}}\right|={\frac {|1-\cos(x)|}{x^{2}}}{\underset {\text{inequality}}{\overset {\text{triangle}}{=}}}{\frac {1+\overbrace {|\cos(x)|} ^{\leq 1}}{x^{2}}}\leq {\frac {2}{x^{2}}}{\overset {x\to \infty }{\to }}0

Using the squeeze theorem, it follows $\lim _{x\to \infty }{\frac {1-\cos(x)}{x^{2}}}=0$ .

Part 8:

\lim _{x\to 0}{\frac {x^{3}-x^{2}}{\tan(x)}}\ {\underset {\text{L.H.}}{\overset {\frac {0}{0}}{=}}}\ \lim _{x\to 0}{\frac {3x^{2}-2x}{1+\tan(x)^{2}}}={\frac {3\cdot 0^{2}-2\cdot 0}{1+\tan(0)^{2}}}={\frac {0}{1+0}}=0

Part 9:

\lim _{x\to 0}{\frac {3^{x}-2^{x}}{\tan(x)}}\ {\underset {\text{L.H.}}{\overset {\frac {0}{0}}{=}}}\ \lim _{x\to 0}{\frac {3^{x}\ln(3)-2^{x}\ln(2)}{1+\tan(x)^{2}}}={\frac {3^{0}\ln(3)-2^{0}\ln(2)}{1+\tan(0)^{2}}}={\frac {\ln(3)-\ln(2)}{1+0}}=\ln \left({\tfrac {3}{2}}\right)

Part 10:

\lim _{x\to 0}{\frac {e^{x}-1-x}{x(e^{x}-1)}}\ {\underset {\text{L.H.}}{\overset {\frac {0}{0}}{=}}}\ \lim _{x\to 0}{\frac {e^{x}-1}{1\cdot (e^{x}-1)+xe^{x}}}\ {\underset {\text{L.H.}}{\overset {\frac {0}{0}}{=}}}\ {\frac {e^{x}}{e^{x}+e^{x}+xe^{x}}}={\frac {e^{0}}{e^{0}+e^{0}+0}}={\frac {1}{2}}

Exercise (L'Hospital 2)

Compute the following limits:

$\lim _{x\to 0+}x^{2}\ln(x^{2})$
$\lim _{x\to 0+}x^{2}\ln(x)^{2}$
$\lim _{x\to 0+}x^{\alpha }\ln(x)^{k}$ for $k\in \mathbb {N} ,\alpha >0$
$\lim _{x\to 1-}\sin(\pi x)\ln(1-x)$
$\lim _{x\to 1}x^{\frac {1}{1-x}}$
$\lim _{n\to \infty }n^{\frac {1}{\sqrt {n}}}$
$\lim _{x\to {\frac {\pi }{2}}-}\cos(x)^{x-{\frac {\pi }{2}}}$
$\lim _{x\to 0}(1+\arctan(x))^{\frac {1}{x}}$
$\lim _{x\to 0}{\frac {1}{x^{2}}}-{\frac {1}{\sin ^{2}(x)}}$
$\lim _{x\to 1}{\frac {a}{1-x^{a}}}-{\frac {b}{1-x^{b}}}$ for $a,b>0$

Solution (L'Hospital 2)

Part 1:

{\begin{aligned}\lim _{x\to 0+}x^{2}\ln(x^{2})&=\lim _{x\to 0}x^{2}\cdot 2\ln(x)\\[0.3em]&=\lim _{x\to 0+}{\frac {2\ln(x)}{\frac {1}{x^{2}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0+}2\ln(x)=-\infty {\text{ and }}\lim _{x\to 0+}{\frac {1}{x^{2}}}=\infty \rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 0+}{\frac {\left(2\ln(x)\right)'}{\left({\frac {1}{x^{2}}}\right)'}}\\[0.3em]&=\lim _{x\to 0+}{\frac {\frac {2}{x}}{-{\frac {2}{x^{3}}}}}\\[0.3em]&=\lim _{x\to 0+}{\frac {-2x^{3}}{2x}}\\[0.3em]&=\lim _{x\to 0+}(-x^{2})\\[0.3em]&=0\end{aligned}}

and since the limit exists, the application of L'Hospital is justified.

Part 2:

{\begin{aligned}\lim _{x\to 0+}x^{2}\ln(x)^{2}&=\lim _{x\to 0+}{\frac {\ln(x)^{2}}{\frac {1}{x^{2}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0+}\ln(x)^{2}=\infty {\text{ and }}\lim _{x\to 0+}{\frac {1}{x^{2}}}=\infty \rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 0+}{\frac {\left(\ln(x)^{2}\right)'}{\left({\frac {1}{x^{2}}}\right)'}}\\[0.3em]&=\lim _{x\to 0+}{\frac {2\ln(x)\cdot {\frac {1}{x}}}{-{\frac {2}{x^{3}}}}}\\[0.3em]&=\lim _{x\to 0+}{\frac {-2x^{3}\ln(x)}{2x}}\\[0.3em]&=\lim _{x\to 0+}-x^{2}\ln(x)\\[0.3em]&=\lim _{x\to 0+}{\frac {-\ln(x)}{\frac {1}{x^{2}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0+}-\ln(x)=\infty {\text{ and }}\lim _{x\to 0+}{\frac {1}{x^{2}}}=\infty \rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 0+}{\frac {-{\frac {1}{x}}}{-{\frac {2}{x^{3}}}}}\\[0.3em]&=\lim _{x\to 0+}({\frac {1}{2}}x^{2})\\[0.3em]&=0\end{aligned}}

and since the limit exists, the application of L'Hospital is justified.

Part 3:

{\begin{aligned}\lim _{x\to 0+}x^{\alpha }\ln(x)^{k}&{\overset {0\cdot (\pm \infty )}{=}}\lim _{x\to 0}{\frac {\ln(x)^{k}}{\frac {1}{x^{\alpha }}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0+}\ln(x)^{k}=\pm \infty {\text{ and }}\lim _{x\to 0+}{\frac {1}{x^{\alpha }}}=\infty \rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 0+}{\frac {k\cdot \ln(x)^{k-1}\cdot {\frac {1}{x}}}{-\alpha {\frac {1}{x^{\alpha +1}}}}}\\[0.3em]&=\lim _{x\to 0+}{\frac {k\cdot \ln(x)^{k-1}}{\frac {-\alpha }{x^{\alpha }}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0+}k\cdot \ln(x)^{k-1}=\pm \infty {\text{ and }}\lim _{x\to 0+}{\frac {-\alpha }{x^{\alpha }}}=\infty \rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 0+}{\frac {k(k-1)\cdot \ln(x)^{k-2}\cdot {\frac {1}{x}}}{\frac {-\alpha \cdot (-\alpha )}{x^{\alpha +1}}}}\\[0.3em]&=\lim _{x\to 0+}{\frac {k(k-1)\cdot \ln(x)^{k-2}}{\frac {(-1)^{2}\alpha ^{2}}{x^{\alpha }}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0+}k(k-1)\cdot \ln(x)^{k-2}=\pm \infty {\text{ and }}\lim _{x\to 0+}{\frac {(-1)^{2}\alpha ^{2}}{x^{\alpha }}}=\infty \rightarrow {\text{L'Hospital further }}(k-2){\text{ times}}\right.}\\[0.3em]&=\lim _{x\to 0+}{\frac {k(k-1)\cdot 2\cdot 1\cdot \ln(x)^{0}}{\frac {(-1)^{k}\alpha ^{k}}{x^{\alpha }}}}\\[0.3em]&=\lim _{x\to 0+}{\frac {k!\cdot 1}{\frac {(-1)^{k}\alpha ^{k}}{x^{\alpha }}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ k!={\text{constant and }}\lim _{x\to 0+}{\frac {(-1)^{k}\alpha ^{k}}{x^{\alpha }}}=\infty \right.}\\[0.3em]&=0\end{aligned}}

and since the limit exists, the application of L'Hospital is justified.

Part 4:

{\begin{aligned}\lim _{x\to 1-}\sin(\pi x)\ln(1-x)&=\lim _{x\to 1-}{\frac {\ln(1-x)}{\frac {1}{\sin(\pi x)}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 1-}\ln(1-x)=-\infty {\text{ and }}\lim _{x\to 1-}{\frac {1}{\sin(\pi x)}}=\infty \rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 1-}{\frac {\left(\ln(1-x)\right)'}{\left({\frac {1}{\sin(\pi x)}}\right)'}}\\[0.3em]&=\lim _{x\to 1-}{\frac {\frac {-1}{1-x}}{\frac {\pi \cos(\pi x)}{\sin(\pi x)^{2}}}}\\[0.3em]&=\lim _{x\to 1-}{\frac {-\sin(\pi x)^{2}}{\pi (1-x)\cos(\pi x)}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 1-}-\sin(\pi x)^{2}=0{\text{ and }}\lim _{x\to 1-}\pi (1-x)\cos(\pi x)=0\rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 1-}{\frac {-2\pi \sin(\pi x)\cos(\pi x)}{-\pi \cos(\pi x)-\pi ^{2}(1-x)\sin(\pi x)}}\\[0.3em]&={\frac {-2\pi \overbrace {\sin(\pi )} ^{=0}\cos(\pi )}{-\pi \underbrace {\cos(\pi )} _{=-1}-\pi ^{2}\cdot 0\cdot \sin(\pi )}}\\[0.3em]&={\frac {0}{\pi }}\\[0.3em]&=0\end{aligned}}

Part 5:

{\begin{aligned}\lim _{x\to 1}x^{\frac {1}{1-x}}&{\overset {1^{\pm \infty }}{=}}\lim _{x\to 1}\exp \left({\frac {\ln(x)}{1-x}}\right)\end{aligned}}

For the expression in the exponent there is now

{\begin{aligned}&\lim _{x\to 1}{\frac {\ln(x)}{1-x}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 1}\ln(x)=0{\text{ and }}\lim _{x\to 1}1-x=0\rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 1}{\frac {\left(\ln(x)\right)'}{\left(1-x\right)'}}\\[0.3em]&=\lim _{x\to 1}{\frac {\frac {1}{x}}{-1}}\\[0.3em]&=\lim _{x\to 1}-{\frac {1}{x}}\\[0.3em]&=-1\end{aligned}}

and since the limit exists, the application of L'Hospital is justified.

By continuity of $x\mapsto \exp(x)$ at $x=-1$ we now get

{\begin{aligned}\lim _{x\to 1}x^{\frac {1}{1-x}}&=\lim _{x\to 1}\exp \left({\frac {\ln(x)}{1-x}}\right)\\[0.3em]&=\exp \left(\lim _{x\to 1}{\frac {\ln(x)}{1-x}}\right)\\[0.3em]&=\exp \left(-1\right)\\[0.3em]&={\frac {1}{e}}\end{aligned}}

Part 6: First we have: If the limit $\lim _{x\to \infty }x^{\frac {1}{\sqrt {x}}}$ exists, then the sequence limit $\lim _{n\to \infty }n^{\frac {1}{\sqrt {n}}}$ also exists.

Further:

\lim _{x\to \infty }x^{\frac {1}{\sqrt {x}}}\ {\overset {{\infty }^{0}}{=}}\ \lim _{x\to \infty }\exp \left({\frac {\ln(x)}{\sqrt {x}}}\right)

For the expression in the exponent there is now

{\begin{aligned}&\lim _{x\to \infty }{\frac {\ln(x)}{\sqrt {x}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to \infty }\ln(x)=\infty {\text{ and }}\lim _{x\to \infty }{\sqrt {x}}=\infty \rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to \infty }{\frac {\left(\ln(x)\right)'}{\left({\sqrt {x}}\right)'}}\\[0.3em]&=\lim _{x\to \infty }{\frac {\frac {1}{x}}{\frac {1}{2{\sqrt {x}}}}}\\[0.3em]&=\lim _{x\to \infty }{\frac {2{\sqrt {x}}}{x}}\\[0.3em]&=\lim _{x\to \infty }{\frac {2}{\sqrt {x}}}\\[0.3em]&=0\end{aligned}}

and since the limit exists, the application of L'Hospital is justified.

By continuity of $x\mapsto \exp(x)$ at $x=0$ we now get

{\begin{aligned}\lim _{x\to \infty }x^{\frac {1}{\sqrt {x}}}&=\lim _{x\to \infty }\exp \left({\frac {\ln(x)}{\sqrt {x}}}\right)\\[0.3em]&=\exp \left(\lim _{x\to \infty }{\frac {\ln(x)}{\sqrt {x}}}\right)\\[0.3em]&=\exp \left(0\right)\\[0.3em]&=1\end{aligned}}

And now, we also have $\lim _{n\to \infty }n^{\frac {1}{\sqrt {n}}}=1$ .

Part 7:

\lim _{x\to {\frac {\pi }{2}}-}\cos(x)^{x-{\frac {\pi }{2}}}\ {\overset {{0}^{0}}{=}}\ \lim _{x\to {\frac {\pi }{2}}-}\exp \left(\ln(\cos(x))(x-{\frac {\pi }{2}})\right)=\lim _{x\to {\frac {\pi }{2}}-}\exp \left({\frac {\ln(\cos(x))}{\frac {1}{x-{\frac {\pi }{2}}}}}\right)

For the expression in the exponent there is now

{\begin{aligned}&\lim _{x\to {\frac {\pi }{2}}-}{\frac {\ln(\cos(x))}{\frac {1}{x-{\frac {\pi }{2}}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to {\frac {\pi }{2}}-}\ln(\underbrace {\cos(x)} _{\to 0+})=-\infty {\text{ and }}\lim _{x\to {\frac {\pi }{2}}-}{\frac {1}{x-{\frac {\pi }{2}}}}=-\infty \rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to {\frac {\pi }{2}}-}{\frac {\left(\ln(\cos(x))\right)'}{\left({\frac {1}{x-{\frac {\pi }{2}}}}\right)'}}\\[0.3em]&=\lim _{x\to {\frac {\pi }{2}}-}{\frac {\frac {-\sin(x)}{\cos(x)}}{-{\frac {1}{(x-{\frac {\pi }{2}})^{2}}}}}\\[0.3em]&=\lim _{x\to {\frac {\pi }{2}}-}{\frac {\sin(x)(x-{\frac {\pi }{2}})^{2}}{\cos(x)}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to {\frac {\pi }{2}}-}\sin(x)(x-{\frac {\pi }{2}})^{2}=0{\text{ and }}\lim _{x\to {\frac {\pi }{2}}-}\cos(x)=0\rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to {\frac {\pi }{2}}-}{\frac {\cos(x)(x-{\frac {\pi }{2}})^{2}+2\sin(x)(x-{\frac {\pi }{2}})}{-\sin(x)}}\\[0.3em]&={\frac {0+0}{-1}}\\[0.3em]&=0\end{aligned}}

and since the limit exists, the application of L'Hospital is justified.

By continuity of $x\mapsto \exp(x)$ at $x=0$ we now get

{\begin{aligned}\lim _{x\to {\frac {\pi }{2}}-}\cos(x)^{(x-{\frac {\pi }{2}})}&=\lim _{x\to {\frac {\pi }{2}}-}\exp \left({\frac {\ln(\cos(x))}{\frac {1}{x-{\frac {\pi }{2}}}}}\right)\\[0.3em]&=\exp \left(\lim _{x\to {\frac {\pi }{2}}-}{\frac {\ln(\cos(x))}{\frac {1}{x-{\frac {\pi }{2}}}}}\right)\\[0.3em]&=\exp \left(0\right)\\[0.3em]&=1\end{aligned}}

Part 8:

\lim _{x\to \infty }\arctan(x)^{\frac {1}{x}}\ {\overset {{\infty }^{0}}{=}}\ \lim _{x\to \infty }\exp \left(\ln(\arctan(x))\cdot {\frac {1}{x}}\right)=\lim _{x\to \infty }\exp \left({\frac {\ln(\arctan(x))}{x}}\right)

For the expression in the exponent there is now

{\begin{aligned}&\lim _{x\to \infty }{\frac {\ln(\arctan(x))}{x}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to \infty }\ln(\underbrace {\arctan(x)} _{\to \infty })=\infty {\text{ and }}\lim _{x\to \infty }x=\infty \rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to \infty }{\frac {\left(\ln(\arctan(x))\right)'}{\left(x\right)'}}\\[0.3em]&=\lim _{x\to \infty }{\frac {\frac {1}{\arctan(x)(1+x^{2})}}{1}}\\[0.3em]&=\lim _{x\to \infty }{\frac {1}{\arctan(x)(1+x^{2})}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to \infty }\arctan(x)(1+x^{2})=\infty \right.}\\[0.3em]&=0\end{aligned}}

and since the limit exists, the application of L'Hospital is justified.

By continuity of $x\mapsto \exp(x)$ at $x=0$ we now get

{\begin{aligned}\lim _{x\to \infty }\arctan(x)^{\frac {1}{x}}&=\lim _{x\to \infty }\exp \left({\frac {\ln(\arctan(x))}{x}}\right)\\[0.3em]&=\exp \left(\lim _{x\to \infty }{\frac {\ln(\arctan(x))}{x}}\right)\\[0.3em]&=\exp \left(0\right)\\[0.3em]&=1\end{aligned}}

Part 9:

{\begin{aligned}&\lim _{x\to 0}{\frac {1}{x^{2}}}-{\frac {1}{\sin ^{2}(x)}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0}{\frac {1}{x^{2}}}=\infty {\text{ and }}\lim _{x\to \infty }{\frac {1}{\sin ^{2}(x)}}=\infty \rightarrow {\text{ Type }}\infty -\infty \rightarrow {\text{re-formulate}}\right.}\\[0.3em]&=\lim _{x\to 0}{\frac {\sin ^{2}(x)-x^{2}}{x^{2}\sin ^{2}(x)}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0}\sin ^{2}(x)-x^{2}=0{\text{ and }}\lim _{x\to 0}x^{2}\sin ^{2}(x)=0\rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 0}{\frac {2\sin(x)\cos(x)-2x}{2x\sin ^{2}(x)+2x^{2}\sin(x)\cos(x)}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0}2\sin(x)\cos(x)-2x=0{\text{ and }}\lim _{x\to 0}2x\sin ^{2}(x)+2x^{2}\sin(x)\cos(x)=0\rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 0}{\frac {2\cos ^{2}(x)-2\sin ^{2}(x)-2}{2\sin ^{2}(x)+\underbrace {4x\sin(x)\cos(x)+4x\sin(x)\cos(x)} _{=8x\sin(x)\cos(x)}+2x^{2}\cos ^{2}(x)-2x^{2}\sin ^{2}(x)}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0}2\cos ^{2}(x)-2\sin ^{2}(x)-2=0{\text{ and }}\lim _{x\to 0}2\sin ^{2}(x)+8x\sin(x)\cos(x)+2x^{2}\cos ^{2}(x)-2x^{2}\sin ^{2}(x)=0\rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 0}{\frac {-4\cos(x)\sin(x)-4\sin(x)\cos(x)}{4\sin(x)\cos(x)+8\sin(x)\cos(x)+8x\cos ^{2}(x)-8x\sin ^{2}(x)+4x\cos ^{2}(x)-4x^{2}\cos(x)\sin(x)-4x\sin ^{2}(x)-4x^{2}\sin(x)\cos(x)}}\\[0.3em]&=\lim _{x\to 0}{\frac {-8\sin(x)\cos(x)}{12\sin(x)\cos(x)+12x\cos ^{2}(x)-12x\sin ^{2}(x)-12x^{2}\sin(x)\cos(x)}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0}-8\sin(x)\cos(x)=0{\text{ and }}\lim _{x\to 0}12\sin(x)\cos(x)+12x\cos ^{2}(x)-12x\sin ^{2}(x)-12x^{2}\sin(x)\cos(x)=0\rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 0}{\frac {-8\cos ^{2}(x)+8\sin ^{2}(x)}{12\cos ^{2}(x)-12\sin ^{2}(x)+12\cos ^{2}(x)-24x\cos(x)\sin(x)-12\sin ^{2}(x)-24x\sin(x)\cos(x)-12x\sin(x)\cos(x)-12x^{2}\cos ^{2}(x)+12x^{2}\sin ^{2}(x)}}\\[0.3em]&=\lim _{x\to 0}{\frac {-8\cos ^{2}(x)+8\sin ^{2}(x)}{24\cos ^{2}(x)-24\sin ^{2}(x)-60x\sin(x)\cos(x)-12x^{2}\cos ^{2}(x)+12x^{2}\sin ^{2}(x)}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0}-8\cos ^{2}(x)=-8{\text{ and }}\lim _{x\to 0}24\cos ^{2}(x)=24\rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&={\frac {-8}{24}}\\[0.3em]&=-{\frac {1}{3}}\end{aligned}}

and since the limit exists, the application of L'Hospital is justified.

Part 10:

{\begin{aligned}&\lim _{x\to 1}{\frac {a}{1-x^{a}}}-{\frac {b}{1-x^{b}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 0}{\frac {a}{1-x^{a}}}=\pm \infty {\text{ and }}\lim _{x\to 1}{\frac {b}{1-x^{b}}}=\pm \infty \rightarrow {\text{ Type }}\pm \infty \pm \infty \rightarrow {\text{re-formulate}}\right.}\\[0.3em]&=\lim _{x\to 1}{\frac {a(1-x^{b})-b(1-x^{a})}{(1-x^{a})(1-x^{b})}}\\[0.3em]&=\lim _{x\to 1}{\frac {a-ax^{b}-b+bx^{a}}{1-x^{a}-x^{b}+x^{a+b}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 1}a-ax^{b}-b+bx^{a}=a-a-b+b=0{\text{ and }}\lim _{x\to 1}1-x^{a}-x^{b}+x^{a+b}=1-1-1+1=0\rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 1}{\frac {-abx^{b-1}+abx^{a-1}}{-ax^{a-1}-bx^{b-1}+(a+b)x^{a+b-1}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ \lim _{x\to 1}-abx^{b-1}+abx^{a-1}=-ab+ab=0{\text{ and }}\lim _{x\to 1}-ax^{a-1}-bx^{b-1}+(a+b)x^{a+b-1}=-a-a+(a+b)=0\rightarrow {\text{L'Hospital}}\right.}\\[0.3em]&=\lim _{x\to 1}{\frac {-ab(b-1)x^{b-2}+ab(a-1)x^{a-2}}{-a(a-1)x^{a-2}-b(b-1)x^{b-2}+(a+b)(a+b-1)x^{a+b-2}}}\\[0.3em]&={\frac {-ab(b-1)+ab(a-1)}{-a(a-1)-b(b-1)+(a+b)(a+b-1)}}\\[0.3em]&={\frac {-ab^{2}+ab+a^{2}b-ab}{-a^{2}+a-b^{2}+b+a^{2}+ab-a+ab+b^{2}-b}}\\[0.3em]&={\frac {-ab^{2}+a^{2}b}{2ab}}\\[0.3em]&={\frac {ab(a-b)}{2ab}}\\[0.3em]&={\frac {a-b}{2}}\end{aligned}}

and since the limit exists, the application of L'Hospital is justified.

Exercise (Differentiability at a point)

Let

f:(-1,\infty )\to \mathbb {R} ,\ f(x)={\begin{cases}{\frac {\ln(1+x)}{x}}&{\text{ for }}x\neq 0,\\1&{\text{ for }}x=0\end{cases}}

Show that $f$ is continuous at zero.
Show that $f$ is differentiable on $\mathbb {R} \setminus \{0\}$ and compute the derivative.
Determine by 1. and 2. the derivative $f'(0)$ .

Solution (Differentiability at a point)

Part 1: By L'Hospital there is

\lim _{x\to 0}f(x)=\lim _{x\to 0}{\frac {\ln(1+x)}{x}}{\underset {\text{l.H.}}{\overset {\tfrac {0}{0}}{=}}}\lim _{x\to 0}{\frac {\frac {1}{1+x}}{1}}=\lim _{x\to 0}{\frac {1}{1+x}}=1=f(0)

So $f$ is continuous at zero.

Part 2: Since $\ln$ , $x\mapsto 1+x$ and $x\mapsto x$ are differentiable on $\mathbb {R} \setminus \{0\}$ , the quotient rule yields that $f$ is differentiable. Further there is for $x\neq 0$ :

f'(x)={\frac {{\frac {1}{1+x}}\cdot x-\ln(1+x)\cdot 1}{x^{2}}}={\frac {\frac {x-(1+x)\ln(1+x)}{1+x}}{x^{2}}}={\frac {x-(1+x)\ln(1+x)}{(1+x)x^{2}}}

Part 3: We use the criterion from the theorem above. There is

\lim _{x\to 0}f'(x)=\lim _{x\to 0}{\frac {x-(1+x)\ln(1+x)}{(1+x)x^{2}}}{\underset {\text{l.H.}}{\overset {\tfrac {0}{0}}{=}}}\lim _{x\to 0}{\frac {1-[\ln(1+x)+1]}{x^{2}+2x(1+x)}}=\lim _{x\to 0}{\frac {-\ln(1+x)}{3x^{2}+2x}}{\underset {\text{l.H.}}{\overset {\tfrac {0}{0}}{=}}}\lim _{x\to 0}{\frac {-{\frac {1}{1+x}}}{6x+2}}=\lim _{x\to 0}{\frac {-1}{(1+x)(6x+2)}}=-{\frac {1}{2}}

Using the criterion, $f$ is differentiable at zero with $f'(0)=-{\tfrac {1}{2}}$ .

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