Continuous extensions
Exercises · Derivatives · with worked solutions ·
Exercise 1
For the following function, determine whether there exist continuous extensions at the endpoints of the domain. Then study the differentiability of such extensions.
\[
f:[0,\pi/2)\rightarrow\R,\ f(x)=(\cos x)^{(\pi/2)-x}
\]
Solution
Setting \(y=\pi/2-x\) we have
\[
\begin{array}{l}\ds\lim_{x\to\pi/2^{-}}f(x)=\lim_{x\to\pi/2^{-}}e^{\left(\frac{\pi}{2}-x\right)\log\cos x}=
\lim_{y\to0^{+}}e^{y\log\sin y}=\lim_{y\to0^{+}}e^{y\log(y+o(y))}=\\
\\
=\lim_{y\to0^{+}}e^{y\log(y(1+o(1))}=\lim_{y\to0^{+}}e^{y\log y+y\log(1+o(1))}=e^{0}=1.\end{array}
\]
We obtain a continuous extension by setting
\[
f(\pi/2)=1.
\]
We examine the limit of the difference quotient:
\[
\begin{array}{l}
\ds\lim_{x\to\pi/2^{-}}\frac{f(x)-f(\pi/2)}{x-\pi/2}=
\lim_{x\to\pi/2^{-}}\frac{e^{\left(\frac{\pi}{2}-x\right)\log\cos x}-1}{x-\pi/2}=\\
\\
= \lim_{x\to\pi/2^{-}}\frac{\left(\frac{\pi}{2}-x\right)\log\cos x}{x-\pi/2}=\lim_{x\to\pi/2^{-}}-\log\cos x=+\infty\end{array}
\]
using the equivalence
\[
e^{t}-1\sim t, \ \ t\to0
\]
with
\[
t=\left(\frac{\pi}{2}-x\right)\log\cos x,\ \ \ x\to\pi/2^{-}.
\]
It follows that \(f\) can be extended continuously at \(x=\pi/2\) , but this extension is not differentiable at \(x=\pi/2\) (vertical tangent).
Exercise 2
For the following function, determine whether there exist continuous extensions at the endpoints of the domain. Then study the differentiability of such extensions.
\[
f:(0,1)\rightarrow\R,\ f(x)=(\cos (\pi x/2))^{\log x}
\]
Solution
We have
\[
\lim_{x\to0}f(x)=\lim_{x\to0}e^{\log x\log\cos (\pi x/2)}=e^{0}=1
\]
since
\[
\begin{array}{l}\ds\lim_{x\to0}\log x\log\cos (\pi x/2)=\lim_{x\to0}\log x\log\left(1-\frac{\pi^{2}}{8}x^{2}+o\left(x^{2}\right)\right)=\\
\\
= \lim_{x\to0}\left(-\frac{\pi^{2}}{8}x^{2}+o\left(x^{2}\right)\right)\log x=\lim_{x\to0}-\frac{\pi^{2}}{8}x^{2}\log x=0.\end{array}
\]
We obtain a continuous extension by setting
\[
f(0)=1.
\]
Solution
We examine the limit of the difference quotient:
\[
\begin{array}{l}
\ds\lim_{x\to0}\frac{f(x)-f(0)}{x}=
\lim_{x\to0}\frac{e^{\log x\log\cos (\pi x/2)}-1}{x}=\\
\\
=\lim_{x\to0}\frac{\log x\log\cos (\pi x/2)}{x}=\lim_{x\to0}-\frac{\pi^{2}}{8}x\log x=0\end{array}
\]
using the equivalences
\[
e^{t}-1\sim t, \ \ t\to0
\]
with
\[
t=\log x\log\cos (\pi x/2),\ \ \ x\to0
\]
and
\[
\log\cos (\pi x/2)\sim -\frac{\pi^{2}}{8}x^{2},\ \ \ x\to0.
\]
It follows that \(f\) can be extended continuously at \(x=0\) , and this extension is differentiable at \(x=0\) with
\[
f'(0)=0.
\]
Solution
We now examine the behavior for \(x\to1\) by setting \(y=x-1\) :
\[
\lim_{x\to1}f(x)=\lim_{x\to1}e^{\log x\log\cos (\pi x/2)}=\lim_{y\to0}e^{\log(1+y)\log(-\sin(\pi y/2))}=e^{0}=1
\]
since
\[
\begin{array}{l}\ds\lim_{y\to0}\log (1+y)\log(-\sin (\pi y/2))=\lim_{y\to0}y\log\left(-\pi y/2+o(y)\right)=\\
\\
=\lim_{y\to0}y\log\left(-\pi y/2(1+o(1))\right)=\\
\\
=\lim_{y\to0}y\log\left(-\pi y/2)+\lim_{y\to0}y\log(1+o(1))\right)=0+0=0.\end{array}
\]
We obtain a continuous extension by setting
\[
f(1)=1.
\]
We examine the limit of the difference quotient:
\[
\begin{array}{l}
\ds\lim_{x\to1}\frac{f(x)-f(1)}{x-1}=
\lim_{x\to1}\frac{e^{\log x\log\cos (\pi x/2)}-1}{x-1}=\lim_{y\to0}\frac{e^{\log(1+y)\log(-\sin(\pi y/2))}-1}{y}=\\
\\
=\lim_{y\to0}\frac{\log(1+y)\log(-\sin(\pi y/2))}{y}=\lim_{y\to0}\log(-\sin(\pi y/2))=-\infty\end{array}
\]
using the equivalences
\[
e^{t}-1\sim t, \ \ t\to0
\]
with
\[
t=\log(1+y)\log(-\sin(\pi y/2)),\ \ \ y\to0
\]
and
\[
\log(1+y)\sim y,\ \ \ y\to0.
\]
It follows that \(f\) can be extended continuously at \(x=1\) , but this extension is not differentiable at \(x=1\) (vertical tangent).
Exercise 3
For the following function, determine whether there exist continuous extensions at the endpoints of the domain. Then study the differentiability of such extensions.
\[
f:(0,1]\rightarrow\R,\ f(x)=\frac{\sqrt{\cos x}-1}{x^{2}}
\]
Solution
We expand the function \(\sqrt{\cos x}\) to order \(3\) with initial point \(x=0\) , using
\[
\cos x=1-\frac{1}{2}x^{2}+o(x^{3})
\]
and
\[
\sqrt{1+y}=1+\frac{1}{2}y-\frac{1}{8}y^{2}+\frac{1}{16}y^{3}+o(y^{3})
\]
with \(y=-(1/2)x^{2}+o(x^{2})\) :
\[
\sqrt{\cos x}=1-\frac{1}{4}x^{2}+o(x^{3}).
\]
It follows that
\[
f(x)=\frac{\sqrt{\cos x}-1}{x^{2}}=-\frac{1}{4}+o(x)
\]
hence \(f\) can be extended continuously at \(x=0\) by setting
\[
f(0)=-\frac{1}{4}
\]
and this extension turns out to be differentiable with
\[
f'(0)=0.
\]
Indeed
\[
\lim_{x\to0}f(x)=\lim_{x\to0}-\frac{1}{4}+o(x)=-\frac{1}{4}
\]
and
\[
\lim_{x\to0}\frac{f(x)-f(0)}{x}=\lim_{x\to0}\frac{-\frac{1}{4}+o(x)+\frac{1}{4}}{x}=\lim_{x\to0}\frac{o(x)}{x}=0.
\]
Exercise 4
For the following function, determine whether there exist continuous extensions at the endpoints of the domain. Then study the differentiability of such extensions.
\[
f:(0,\pi/2]\rightarrow\R,\ f(x)=(1-\cos x)^{\log(1+x)}
\]
Solution
We have
\[
\lim_{x\to0}f(x)=\lim_{x\to0}e^{\log(1+x)\log(1-\cos x)}=e^{0}=1
\]
since
\[
\begin{array}{l}\ds\lim_{x\to0}\log(1+x)\log(1-\cos x)=\lim_{x\to0} x\log\left(\frac{1}{2}x^{2}+o\left(x^{2}\right)\right)=\\
\\
= \lim_{x\to0}x\log\left(\frac{1}{2}x^{2}(1+o(1)\right)=\\
\\
= \lim_{x\to0}x\log\left(\frac{1}{2}x^{2}\right)+\lim_{x\to0}x\log(1+o(1))=0+0=0.\end{array}
\]
We obtain a continuous extension by setting
\[
f(0)=1.
\]
We examine the limit of the difference quotient:
\[
\begin{array}{l}
\ds\lim_{x\to0}\frac{f(x)-f(0)}{x}=
\lim_{x\to0}\frac{e^{\log(1+x)\log(1-\cos x)}-1}{x}=\\
\\
= \lim_{x\to0}\frac{\log(1+x)\log(1-\cos x)}{x}=\lim_{x\to0}\log(1-\cos x)=-\infty\end{array}
\]
using the equivalence
\[
e^{t}-1\sim t, \ \ t\to0
\]
with
\[
t=\log(1+x)\log(1-\cos x),\ \ \ x\to0.
\]
It follows that \(f\) can be extended continuously at \(x=0\) , but this extension is not differentiable at \(x=0\) (vertical tangent).