(Created page with "== 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...") |
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** know that $f(a) = g(a) = 0$, so have | ** know that $f(a) = g(a) = 0$, so have | ||

* $\lim\limits_{x \to a} \cfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \cfrac{f'(a)\, (x - a) + \ ...}{g'(a)\, (x - a) + \ ...}$ | * $\lim\limits_{x \to a} \cfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \cfrac{f'(a)\, (x - a) + \ ...}{g'(a)\, (x - a) + \ ...}$ | ||

− | * can factor $(x - a)$ out, so we have: | + | * can factor $(x - a)$ out, so now we have: |

* $\lim\limits_{x \to a} \cfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \cfrac{f'(a) + \ ...}{g'(a) + \ ...}$ | * $\lim\limits_{x \to a} \cfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \cfrac{f'(a) + \ ...}{g'(a) + \ ...}$ | ||

* the leading order terms are $f'(a)$ and $g'(a)$, and the rest vanish under the limit | * the leading order terms are $f'(a)$ and $g'(a)$, and the rest vanish under the limit | ||

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* so the limit is 0 | * so the limit is 0 | ||

− | To say that one function grows faster than other, we can use the [[ | + | To say that one function grows faster than other, we can use the [[Orders of Growth|Big-O notation]] |

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

- 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:

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

- 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

See http://calculus.seas.upenn.edu/?n=Main.LHopitalsRule