Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

6.5: L'Hopital's Rule

  • Page ID
    22675
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    The following result is one case of \(l^{\prime}\) l'Hópital's rule.

    Theorem \(\PageIndex{1}\)

    Suppose \(a, b \in \mathbb{R}, f\) and \(g\) are differentiable on \((a, b), g^{\prime}(x) \neq 0\) for all \(x \in(a, b),\) and

    \[\lim _{x \rightarrow a^{+}} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lambda .\]

    If \(\lim _{x \rightarrow a^{+}} f(x)=0\) and \(\lim _{x \rightarrow a^{+}} g(x)=0,\) then

    \[\lim _{x \rightarrow a^{+}} \frac{f(x)}{g(x)}=\lambda .\]

    Proof

    Given \(\epsilon>0,\) there exists \(\delta>0\) such that

    \[\lambda-\frac{\epsilon}{2}<\frac{f^{\prime}(x)}{g^{\prime}(x)}<\lambda+\frac{\epsilon}{2}\]

    whenever \(x \in(a, a+\delta) .\) Now, by the Generalized Mean Value Theorem, for any \(x\) and \(y\) with \(a<x<y<a+\delta,\) there exists a point \(c \in(x, y)\) such that

    \[\frac{f(y)-f(x)}{g(y)-g(x)}=\frac{f^{\prime}(c)}{g^{\prime}(c)}.\]

    Hence

    \[\lambda-\frac{\epsilon}{2}<\frac{f(y)-f(x)}{g(y)-g(x)}<\lambda+\frac{\epsilon}{2}.\]

    Now

    \[\lim _{x \rightarrow a^{+}} \frac{f(y)-f(x)}{g(y)-g(x)}=\frac{f(y)}{g(y)}\]

    and so we have

    \[\lambda-\epsilon<\lambda-\frac{\epsilon}{2} \leq \frac{f(y)}{g(y)} \leq \lambda+\frac{\epsilon}{2}<\lambda+\epsilon\]

    for any \(y \in(a, a+\delta) .\) Hence

    \[\lim _{x \rightarrow a^{+}} \frac{f(x)}{g(x)}=\lambda .\]

    Q.E.D.

    Exercise \(\PageIndex{1}\)

    Use l'Hôpital's rule to compute

    \[\lim _{x \rightarrow 0^{+}} \frac{\sqrt{1+x}-1}{x}. \nonumber\]

    Exercise \(\PageIndex{2}\)

    Suppose \(a, b \in \mathbb{R}, f\) and \(g\) are differentiable on \((a, b), g^{\prime}(x) \neq 0\) for all \(x \in(a, b),\) and

    \[\lim _{x \rightarrow b^{-}} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lambda . \nonumber\]

    Show that if \(\lim _{x \rightarrow b^{-}} f(x)=0\) and \(\lim _{x \rightarrow b^{-}} g(x)=0,\) then

    \[\lim _{x \rightarrow b^{-}} \frac{f(x)}{g(x)}=\lambda . \nonumber\]

    • Was this article helpful?