5.3.E: Problems on \(L^{\prime}\) Hôpital's Rule
Elementary differentiation formulas are assumed known.
Complete the proof of L'Hôpital's rule. Verify that the differentiability assumption may be replaced by continuity plus existence of finite or infinite (but not both together infinite) derivatives \(f^{\prime}\) and \(g^{\prime}\) on \(G_{\neg p}\) (same proof).
Show that the rule fails for complex functions. See, however, Problems \(3,\) \(7,\) and \(8 .\)
[Hint: Take \(p=0\) with
\[
f(x)=x \text { and } g(x)=x+x^{2} e^{i / x^{2}}=x+x^{2}\left(\cos \frac{1}{x^{2}}+i \cdot \sin \frac{1}{x^{2}}\right) .
\]
Then
\[
\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=1, \text { though } \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lim _{x \rightarrow 0} \frac{1}{g^{\prime}(x)}=0 .
\]
Indeed, \(g^{\prime}(x)-1=(2 x-2 i / x) e^{i / x^{2}} .\) (Verify!) Hence
\[
\left.\left|g^{\prime}(x)\right|+1 \geq|2 x-2 i / x| \quad \text { (for }\left|e^{i / x^{2}}\right|=1\right) ,
\]
so
\[
\left|g^{\prime}(x)\right| \geq-1+\frac{2}{x} . \quad(\text { Why? })
\]
Deduce that
\[
\left.\left|\frac{1}{g^{\prime}(x)}\right| \leq\left|\frac{x}{2-x}\right| \rightarrow 0 .\right]
\]
Prove the "simplified rule of \(L^{\prime}\) Hôpital" for real or complex functions \(\text { (also for vector-valued } f \text { and scalar-valued } g) :\) If \(f\) and \(g\) are differentiable at \(p,\) with \(g^{\prime}(p) \neq 0\) and \(f(p)=g(p)=0\), then
\[
\lim _{x \rightarrow p} \frac{f(x)}{g(x)}=\frac{f^{\prime}(p)}{g^{\prime}(p)} .
\]
[Hint:
\[
\frac{f(x)}{g(x)}=\frac{f(x)-f(p)}{g(x)-g(p)}=\frac{\Delta f}{\Delta x} / \frac{\Delta g}{\Delta x} \rightarrow \frac{f^{\prime}(p)}{g^{\prime}(p)} .
\]
Why does \(\lim _{x \rightarrow+\infty} \frac{f(x)}{g(x)}\) not exist, though \(\lim _{x \rightarrow+\infty} \frac{f^{\prime}(x)}{g^{\prime}(x)}\) does, in the following example? Verify and explain.
\[
f(x)=e^{-2 x}(\cos x+2 \sin x), \quad g(x)=e^{-x}(\cos x+\sin x) .
\]
[Hint: \(g^{\prime}\) vanishes many times in each \(G_{+\infty} .\) Use the Darboux property for the \(\text { proof. }]\)
Find \(\lim _{x \rightarrow 0^{+}} \frac{e^{-1 / x}}{x}\).
\(\left.\text { [Hint: Substitute } z=\frac{1}{x} \rightarrow+\infty . \text { Then use the rule. }\right]\)
Verify that the assumptions of L'Hôpital's rule hold, and find the following limits.
(a) \(\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\ln (e-x)+x-1}\);
(b) \(\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}\);
(c) \(\lim _{x \rightarrow 0} \frac{(1+x)^{1 / x}-e}{x}\);
(d) \(\lim _{x \rightarrow 0^{+}}\left(x^{q} \ln x\right), q>0\);
(e) \(\lim _{x \rightarrow+\infty}\left(x^{-q} \ln x\right), q>0\);
(f) \(\lim _{x \rightarrow 0^{+}} x^{x}\);
(g) \(\lim _{x \rightarrow+\infty}\left(x^{q} a^{-x}\right), a>1, q>0\);
(h) \(\lim _{x \rightarrow 0}\left(\frac{1}{x^{2}}-\operatorname{cotan}^{2} x\right)\);
(i) \(\lim _{x \rightarrow+\infty}\left(\frac{\pi}{2}-\arctan x\right)^{1 / \ln x}\);
(j) \(\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{1 /(1-\cos x)}\).
Prove L'Hôpital's rule for \(f : E^{1} \rightarrow E^{n}(C)\) and \(g : E^{1} \rightarrow E^{1},\) with
\[
\lim _{k \rightarrow p}|f(x)|=0=\lim _{x \rightarrow p}|g(x)|, p \in E^{*} \text { and } r \in E^{n} ,
\]
leaving the other assumptions unchanged.
\(\left.\left.\text { [Hint: Apply the rule to the components of } \frac{f}{g} \text { (respectively, to }\left(\frac{f}{g}\right)_{\mathrm{re}} \text { and }\left(\frac{f}{g}\right)_{\mathrm{im}}\right) .\right]\)
Let \(f\) and \(g\) be complex and differentiable on \(G_{\neg p}, p \in E^{1} .\) Let
\[
\lim _{x \rightarrow p} f(x)=\lim _{x \rightarrow p} g(x)=0, \lim _{x \rightarrow p} f^{\prime}(x)=q, \text { and } \lim _{x \rightarrow p} g^{\prime}(x)=r \neq 0 .
\]
Prove that \(\lim _{x \rightarrow p} \frac{f(x)}{g(x)}=\frac{q}{r}\).
[Hint:
\[
\frac{f(x)}{g(x)}=\frac{f(x)}{x-p} / \frac{g(x)}{x-p} .
\]
Apply Problem 7 to find
\[
\lim _{x \rightarrow p} \frac{f(x)}{x-p} \text { and } \lim _{x \rightarrow p} \frac{g(x)}{x-p} .]
\]
Do Problem 8 for \(f : E^{1} \rightarrow C^{n}\) and \(g : E^{1} \rightarrow C\).