L’Hôpital’s rule

Key results

L’Hôpital’s rule

$$\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}$$$\underset{x\to a}{\lim }\frac{f(x)}{g(x)}$ $=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

When $f(x)$ and $g(x)$ both tend to 0, or both to $\pm\infty$, $f'(x)$ and $g'(x)$ exist with $g'(x) \neq 0$ for $x$ near $a$, except possibly at $a$ itself, and the limit on the right exists or is $\pm\infty$; it holds too for one-sided limits and as $x \to \pm\infty$.

Products

$$f(x)\,g(x) = \frac{f(x)}{1/g(x)} = \frac{g(x)}{1/f(x)}$$$f(x)g(x)$ $=\frac{f(x)}{1/g(x)}$ $=\frac{g(x)}{1/f(x)}$

Wherever $f(x) \neq 0$ and $g(x) \neq 0$; when $f \to 0$ and $g \to \pm\infty$, the first quotient is a $\tfrac{0}{0}$ form and the second an $\tfrac{\infty}{\infty}$ form.

Variable powers

$$\lim_{x \to a} f(x)^{g(x)} = e^{L}, \qquad L = \lim_{x \to a} g(x)\log f(x)$$$\underset{x\to a}{\lim }f(x{)}^{g(x)}$ $=e^L,$$L$ $=\underset{x\to a}{\lim }g(x)\log f(x)$

When $f(x) > 0$ for $x$ near $a$ and $L$ exists as a real number; this settles the forms $1^{\infty}$, $0^{0}$ and $\infty^{0}$.

Growth rates

$$\lim_{x \to \infty}\frac{x^{p}}{e^{x}} = 0, \qquad \lim_{x \to \infty}\frac{\log x}{x^{p}} = 0$$$\underset{x\to \infty }{\lim }\frac{x^p}{e^x}$ $=0,$$\underset{x\to \infty }{\lim }\frac{\log x}{x^p}$ $=0$

For any constant $p > 0$.

Continuous compounding

$$\lim_{x \to \infty}\Bigl(1 + \frac{a}{x}\Bigr)^{x} = e^{a}$$$\underset{x\to \infty }{\lim }(1+\frac{a}{x}{)}^x$ $=e^a$

For any constant $a$.

·Read the theory·UBC Math 100 §6 L’Hôpital’s rule and indeterminate forms·CLP-1 §3.7 L’Hôpital’s rule and indeterminate forms