# L'Hopital Rule

• Aug 8th 2008, 01:14 PM
particlejohn
L'Hopital Rule
Prove the following: Let $a be real numbers, let $f: [a,b] \to \bold{R}$ and $g: [a,b] \to \bold{R}$ be functions which are differentiable on $[a,b]$. Suppose that $f(a) = g(a) = 0$, that $g'$ is nonzero on $[a,b]$, and $\lim_{x \to a: x \in (a,b]} \frac{f'(x)}{g'(x)}$ exists and equals $L$. Then $g(x) \neq 0$ for all $x \in (a,b]$, and $\lim_{x \to a: x \in (a,b]} \frac{f(x)}{g(x)}$ exists and equals $L$.

Suppose that $g(x) = 0$ for some $x \in (a,b]$. Since $g(a) = 0$, then by Rolle's theorem we can find a $y \in (a,b]$ such that $g'(y) = 0$. But this contradicts the fact that $g'$ is nonzero on $[a,b]$. Hence $g(x) \neq 0$ for all $x \in (a,b]$.

Now to show that $\lim_{x \to a: x \in (a,b]} \frac{f'(x)}{g'(x)}$ exists and equals $L$, we can show that $\lim_{n \to \infty} \frac{f(x_{n})}{g(x_{n})} = L$ for any sequence $(x_{n})_{n=1}^{\infty}$ which converges to $x$. How would I show this? I am guessing that I have to create a new function, and somehow manipulate it so that $\frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)}$. And this is equivalent to showing that $\frac{f(x_{n})}{g(x_{n})} = \frac{f'(x_{n})}{g'(x_{n})}$.
• Aug 8th 2008, 03:01 PM
wingless
The mean value theorem says that there is an x in the interval (a,b) such that $g'(x) [f(b)-f(a)] = f'(x)[g(b)-g(a)]$.

Now, we know that g' is nonzero (you assumed it), and if g(a) $\neq$ g(b), we can write $\frac{f'(x)}{g'(x)} = \frac{f(b)-f(a)}{g(b)-g(a)}$.

As we know f(a) = g(a) = 0,

$\frac{f'(x)}{g'(x)} = \frac{f(b)}{g(b)}$

If we take the limit of both sides, then x and b will approach the same point.

$\lim_{x\to x_0}\frac{f'(x)}{g'(x)} = \lim_{b\to b_0}\frac{f(b)}{g(b)} = \lim_{x\to x_0}\frac{f(x)}{g(x)}$

$\lim_{x\to x_0}\frac{f(x)}{g(x)} = \lim_{x\to x_0}\frac{f'(x)}{g'(x)}$
• Aug 8th 2008, 03:24 PM
particlejohn
You could use Rolle's Theorem also right? In terms of sequences you can define a function $z_{n}: [a, x_{n}] \to \bold{R}$ by $z_{n}(x) = f(x)g(x_{n}) - g(x)f(x_{n})$. So this is continuous on the interval, and is zero at the endpoints. And $z'_{n}(x) = f'(x)g(x_{n})- g'(x)f(x_{n})$. Using Rolle's Theorem we get $\frac{f(x_{n})}{g(x_{n})} = \frac{f'(y_{n})}{g'(y_{n})}$. $x_n \to a$ as $n \to \infty$ and so $y_n \in (a, x_n) \to a$ as $n \to \infty$ by Squeeze Theorem.

And so $\frac{f'(x_{n})}{g'(x_{n})} \to L, \frac{f'(y_{n})}{g'(y_{n})} \to L \implies \frac{f(x_{n})}{g(x_{n})} \to L$.
• Aug 8th 2008, 04:00 PM
wingless
Quote:

Originally Posted by particlejohn
You could use Rolle's Theorem also right? In terms of sequences you can define a function $z_{n}: [a, x_{n}] \to \bold{R}$ by $z_{n}(x) = f(x)g(x_{n}) - g(x)f(x_{n})$. So this is continuous on the interval, and is zero at the endpoints. And $z'_{n}(x) = f'(x)g(x_{n})- g'(x)f(x_{n})$. Using Rolle's Theorem we get $\frac{f(x_{n})}{g(x_{n})} = \frac{f'(y_{n})}{g'(y_{n})}$. $x_n \to a$ as $n \to \infty$ and so $y_n \in (a, x_n) \to a$ as $n \to \infty$ by Squeeze Theorem.

And so $\frac{f'(x_{n})}{g'(x_{n})} \to L, \frac{f'(y_{n})}{g'(y_{n})} \to L \implies \frac{f(x_{n})}{g(x_{n})} \to L$.

But what you did here was proving Mean Value Theorem using Rolle's Theorem =)
(except you used sequences instead of function. see here)