6.5: L'Hopital's Rule
- Page ID
- 22675
The following result is one case of \(l^{\prime}\) l'Hópital's rule.
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.
Use l'Hôpital's rule to compute
\[\lim _{x \rightarrow 0^{+}} \frac{\sqrt{1+x}-1}{x}. \nonumber\]
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
\[\lim _{x \rightarrow b^{-}} \frac{f(x)}{g(x)}=\lambda . \nonumber\]