Skip to main content
Mathematics LibreTexts

5.3.E: Problems on \(L^{\prime}\) Hôpital's Rule

  • Page ID
    23754
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Elementary differentiation formulas are assumed known.

    Exercise \(\PageIndex{1}\)

    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).

    Exercise \(\PageIndex{2}\)

    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
    \[
    \left.\left|\frac{1}{g^{\prime}(x)}\right| \leq\left|\frac{x}{2-x}\right| \rightarrow 0 .\right]
    \]

    Exercise \(\PageIndex{3}\)

    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:
    \[
    \frac{f(x)}{g(x)}=\frac{f(x)-f(p)}{g(x)-g(p)}=\frac{\Delta f}{\Delta x} / \frac{\Delta g}{\Delta x} \rightarrow \frac{f^{\prime}(p)}{g^{\prime}(p)} .
    \]

    Exercise \(\PageIndex{4}\)

    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 \(\text { proof. }]\)

    Exercise \(\PageIndex{5}\)

    Find \(\lim _{x \rightarrow 0^{+}} \frac{e^{-1 / x}}{x}\).
    \(\left.\text { [Hint: Substitute } z=\frac{1}{x} \rightarrow+\infty . \text { Then use the rule. }\right]\)

    Exercise \(\PageIndex{6}\)

    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) \(\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{1 /(1-\cos x)}\).

    Exercise \(\PageIndex{7}\)

    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.
    \(\left.\left.\text { [Hint: Apply the rule to the components of } \frac{f}{g} \text { (respectively, to }\left(\frac{f}{g}\right)_{\mathrm{re}} \text { and }\left(\frac{f}{g}\right)_{\mathrm{im}}\right) .\right]\)

    Exercise \(\PageIndex{8}\)

    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
    \[
    \lim _{x \rightarrow p} \frac{f(x)}{x-p} \text { and } \lim _{x \rightarrow p} \frac{g(x)}{x-p} .]
    \]

    Exercise \(\PageIndex{*9}\)

    Do Problem 8 for \(f : E^{1} \rightarrow C^{n}\) and \(g : E^{1} \rightarrow C\).


    5.3.E: Problems on \(L^{\prime}\) Hôpital's Rule is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?