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


Other Cases

Links

Sources

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