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L'Hôpital's Rule

The rule has this form: $$\lim_{x \to a} \cfrac{f(x)}{g(x)} = \lim_{x \to a} \cfrac{f'(x)}{g'(x)}$$

$0/0$ case

• suppose $\lim\limits_{x \to a} \cfrac{f(x)}{g(x)} = \cfrac{0}{0}$
• i.e. $f(a) = 0$ and $g(a) = 0$
• if $f(x)$ and $g(x)$ are continuous
• then the rule is $$\lim_{x \to a} \cfrac{f(x)}{g(x)} = \lim_{x \to a} \cfrac{f'(x)}{g'(x)}$$

Can show this using Taylor Expansion about $x = a$:

• $\lim\limits_{x \to a} \cfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \cfrac{f(a) + f'(a)\, (x - a) + \ ...}{g(a) + g'(a)\, (x - a) + \ ...}$
• know that $f(a) = g(a) = 0$, so have
• $\lim\limits_{x \to a} \cfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \cfrac{f'(a)\, (x - a) + \ ...}{g'(a)\, (x - a) + \ ...}$
• can factor $(x - a)$ out, so now we have:
• $\lim\limits_{x \to a} \cfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \cfrac{f'(a) + \ ...}{g'(a) + \ ...}$
• the leading order terms are $f'(a)$ and $g'(a)$, and the rest vanish under the limit

Examples:

• $\lim\limits_{x \to 0} \cfrac{\sin x}{x} = \lim\limits_{x \to 0} \cfrac{\cos x}{1} = 1$
• $\lim\limits_{x \to 0} \cfrac{1 - \cos x}{x} = \lim\limits_{x \to 0} \cfrac{\sin x}{1} = 0$

$\infty / \infty$ case

• suppose $\lim\limits_{x \to a} \cfrac{f(x)}{g(x)} = \cfrac{\infty}{\infty}$
• i.e. $f(a) = \infty$ and $g(a) = \infty$
• if $f(x)$ and $g(x)$ are continuous
• then the rule is $$\lim_{x \to a} \cfrac{f(x)}{g(x)} = \lim_{x \to a} \cfrac{f'(x)}{g'(x)}$$

Example:

• $\lim\limits_{x \to \infty} \cfrac{\ln x}{\sqrt{x}}$
• both go to $\infty$, but the rate at which they approach $\infty$ is different
• by taking the derivative, we can see which one grows faster
• is this case, $\sqrt{x}$ dominates $\ln x$: it grows much faster
• so the limit is 0

To say that one function grows faster than other, we can use the Big-O notation