# Thread: Proving L'Hopital's (Bernoulli's) Rule

1. ## Proving L'Hopital's (Bernoulli's) Rule

I wish to prove the following.

"Let $f(x), g(x)$ be represented by Taylor series at $x = a$, where $f(a) = g(a) = 0$, and let the first non-vanishing derivatives at $a$ be $f^{(M)} (a)$ and $g^{(N)}(a)$, where $M \geq N \geq 1$. Then

$\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f^{(M)}(a)}{g^{(N)}(a)}$"

So here's how I've proceeded so far:

$f(x) = A_0 + A_1 (x - a) + A_2 (x - a)^2 + \ldots$

If we let $x = a$, then $A_0 = f(a)$

Differentiating wrt $x$:

$f'(x) = 1 \cdot A_1 + 2 \cdot A_2 (x - a)^1 + \ldots$

Again, letting $x = a$, then $A_1 = \frac{1}{1!}f'(a)$

And so on, so that:

$f(x) = f(a) + \frac{1}{1!} f^{(1)}(a) (x - a) + \frac{1}{2!} f^{(2)} (a) (x - a)^2 + \ldots$

and similarly,

$g(x) = g(a) + \frac{1}{1!} g^{(1)} (a) (x - a) + \frac{1}{2!} g^{(2)} (a) (x - a)^2 + \ldots$

So:

$\frac{f(x)}{g(x)} = \frac{ f(a) + \frac{1}{1!} f^{(1)} (a) ( x - a ) + \frac{1}{2!} f^{(2)} (a) (x - a)^2 + \ldots }{g(a) + \frac{1}{1!} g^{(1)} (a) (x - a) + \frac{1}{2!} g^{(2)} (a) (x - a)^2 + \ldots}$

Recalling that $f^{(M)} (a)$ and $g^{(N)} (a)$ are the first non-vanishing derivatives at $a$. Then:

$\frac{f(x)}{g(x)} = \frac{\frac{1}{M!} f^{(M)} (a) (x - a)^M + \textrm{higher powers}}{\frac{1}{N!} g^{(N)} (a) (x - a)^N + \textrm{higher powers}}$

Recalling that $M \geq N \geq 1$, then:

$\frac{f(x)}{g(x)} = \frac{\frac{1}{M!} f^{(M)} (a) (x - a)^{M-N} + \textrm{higher powers}}{\frac{1}{N!} g^{(N)} (a) + \textrm{higher powers}}$

But unless $M = N$, as $x \to a$, $(x - a)^{M - N} \to 0$ and so $\frac{f(x)}{g(x)} \to 0$, which clearly is not the desired result.

Where am I going wrong?

2. Originally Posted by Harry1W
I wish to prove the following.

"Let $f(x), g(x)$ be represented by Taylor series at $x = a$, where $f(a) = g(a) = 0$, and let the first non-vanishing derivatives at $a$ be $f^{(M)} (a)$ and $g^{(N)}(a)$, where $M \geq N \geq 1$. Then

$\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f^{(M)}(a)}{g^{(N)}(a)}$"

So here's how I've proceeded so far:

...
[SNIP]
...

Recalling that $M \geq N \geq 1$, then:

$\frac{f(x)}{g(x)} = \frac{\frac{1}{M!} f^{(M)} (a) (x - a)^{M-N} + \textrm{higher powers}}{\frac{1}{N!} g^{(N)} (a) + \textrm{higher powers}}$

But unless $M = N$, as $x \to a$, $(x - a)^{M - N} \to 0$ and so $\frac{f(x)}{g(x)} \to 0$, which clearly is not the desired result.

Where am I going wrong?
You're not going wrong. The "desired result" is not true unless M=N. As your analysis correctly shows, the limit will be zero if M>N.