L'Hôpital's Rule

Evaluation

$\frac{0}{0}$ or $\frac{\infty}{\infty}$

Apply L'Hôpital's Rule, several times if needed.
If the derivative loops, L'Hôpital's Rule cannot be used.

[!example]- Evaluate the limit $lim_{x \to 0^{+}} \frac{\ln x}{1 + (\ln x)^{2}}$ .
$\begin{aligned} lim_{x \to 0^{+}} \frac{\ln x}{1 + (\ln x)^{2}}, x = 0, - \frac{\infty}{\infty} \\ = & lim_{x \to 0^{+}} \frac{1}{2 \ln x} \\ = & 0 \end{aligned}$

$\infty - \infty$

[!example]- Evaluate the limit $lim_{x \to \infty} (\ln x - \sqrt{x - 1})$ .
$\begin{aligned} lim_{x \to \infty} (\ln x - \sqrt{x - 1}), x \to \infty, \infty - \infty \\ = & lim_{x \to \infty} (\sqrt{x - 1} (\frac{\ln x}{\sqrt{x - 1}} - 1)) \end{aligned}$
On the side:
$\begin{aligned} lim_{x \to \infty} \frac{\ln x}{\sqrt{x - 1}} & = lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{2 \sqrt{x - 1}}} \\ = lim_{x \to \infty} \frac{1}{x} \cdot 2 \sqrt{x - 1} \\ = lim_{x \to \infty} \frac{2 \sqrt{x - 1}}{x} \\ = lim_{x \to \infty} \frac{2 \sqrt{x} \sqrt{1 - \frac{1}{x}}}{x} \\ = lim_{x \to \infty} \frac{2 \sqrt{1 - \frac{1}{x}}}{\sqrt{x}} \\ = 0 \end{aligned}$
Back to the limit,
$\begin{aligned} lim_{x \to \infty} (\sqrt{x - 1} (\frac{\ln x}{\sqrt{x - 1}} - 1)) & = lim_{x \to \infty} \sqrt{x - 1} (lim_{x \to \infty} \frac{\ln x}{\sqrt{x - 1}} - lim_{x \to \infty} 1) \\ = \infty (0 - 1) \\ = - \infty \end{aligned}$

Contributors

Ajitani Hifumi

Changelog

Last edited 26 days ago

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L'Hôpital's Rule ​

Evaluation ​

00 or ∞∞ ​

∞−∞ ​

Contributors

Changelog

L'Hôpital's Rule

Evaluation

$\frac{0}{0}$ or $\frac{\infty}{\infty}$

$\infty - \infty$