# L'Hôpital's Rule

L'Hopital's Rule is a theorem dealing with limits that is very important to calculus.

## Theorem

The theorem states that for real functions $f(x),g(x)$, if $\lim f(x),g(x)\in \{0,\pm \infty\}$ $$\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}$$ Note that this implies that $$\lim\frac{f(x)}{g(x)}=\lim\frac{f^{(n)}(x)}{g^{(n)}(x)}=\lim\frac{f^{(-n)}(x)}{g^{(-n)}(x)}$$

## Proof

One can prove using linear approximation: The definition of a derivative is $f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$ which can be rewritten as $f'(x) = \frac{f(x+h)-f(x)}{h} + \eta(h)$. Just so all of us know $\eta(h)$ is a function that is both continuous and has a limit of $0$ as the $h$ in the derivative function approaches $0$. After multiplying the equation above by $h$, we get $f(x+h) = f(x) +f'(x)h+h\cdot \eta(h)$.

We have already assumed by the hypothesis that the derivative equals zero. Hence, we can rewrite the function as $\frac{f(x_0+h)}{g(x_0+h)}=\frac{f'(x_0)h + h\cdot \eta(h)}{g'(x_0)h+h\cdot \epsilon(h)}$, which would hence prove our lemma for L'Hospital's rule.

Text explanation:

Let $z(x) = \frac{f(x)}{g(x)}$, where $f(x)$ and $g(x)$ are both nonzero functions with value $0$ at $x = a$.

(For example, $g(x) = \cos\left(\frac{\pi}{2} x\right)$, $f(x) = 1-x$, and $a = 1$.)

Note that the points surrounding $z(a)$ aren't approaching infinity, as a function like $f(x) = 1/x-1$ might at $f(a)$.

The points infinitely close to $z(a)$ will be equal to $\lim_{b\to 0} \frac{f(a+b)}{g(a+b)}$.

Note that $\lim_{b\to 0} f(a+b)$ and $\lim_{b\to 0} g(a+b)$ are equal to $f'(a)$ and $g'(a)$.

As a recap, this means that the points approaching $\frac{f(a)}{g(a)}$, where $a$ is a number such that $f(a)$ and $g(a)$ are both equal to $0$, are going to approach $\frac{f'(x)}{g'(x)}$.

## Problems

### Introductory

• Evaluate the limit $\lim_{x\rightarrow3}\frac{x^{2}-4x+3}{x^{2}-9}$ (weblog_entry.php?t=168186 Source)