# [SOLVED] Limit when Hopital's rule fails

• Jul 3rd 2008, 01:41 AM
flowtsen
[SOLVED] Limit when Hopital's rule fails
I have a function $\displaystyle f(x) = g^{-1}(h(x))$ and I'm looking for $\displaystyle \lim_{x \rightarrow x_0^+} f'(x)$

I know the following about the functions g and h:
• $\displaystyle h(x_0)=0$ and $\displaystyle g(0) = 0$ so $\displaystyle f(x_0)=0$
• $\displaystyle g \geq 0, h \geq 0$
• $\displaystyle g'(0)=0, h'(x_0)=0$
• $\displaystyle g' > 0$ for $\displaystyle x \neq x_0$, $\displaystyle h' > 0$ for $\displaystyle x<x_0$ and $\displaystyle h'<0$ for $\displaystyle x>x_0$
• $\displaystyle g'' > 0, h'' > 0$ and finite at $\displaystyle x_0$
• $\displaystyle h''' = 0, g'''$ finite at $\displaystyle x_0$
We have $\displaystyle f'(x)= \frac{h'(x)}{g'(f(x))}$. At $\displaystyle x_0$, both numerator and denominator are 0. Applying l'Hopital's rule, we obtain

$\displaystyle \lim_{x \rightarrow x_0^+} f'(x) = \lim_{x \rightarrow x_0^+} \frac{h''(x)}{f'(x) g''(f(x))} = \lim_{x \rightarrow x_0^+} \frac{h''(x) g'(f(x))}{h'(x) g''(f(x))}$.

Applying l'Hopital's rule again, we're back to where we started:

$\displaystyle \lim_{x \rightarrow x_0^+} \frac{h'''(x) g'(f(x))+h''(x) f'(x) g''(f(x))}{h''(x) g''(f(x)) + h'(x) f'(x) g'''(f(x))} = \lim_{x \rightarrow x_0^+} f'(x).$

There's probably a simpler way of presenting this problem, sorry about that... If anyone knows how this kind of circularity problem is called and/or how to deal with it, I will be very grateful!
• Jul 3rd 2008, 01:49 AM
Moo
Hello,

Quote:

$\displaystyle \lim_{x \rightarrow x_0^+} f'(x) = \lim_{x \rightarrow x_0^+} \frac{h''(x)}{f'(x) g''(f(x))}$
When you're here, why don't you multiply both sides by f'(x) ? (if I'm correct, it's not important whether f'(x)=0 or not).

$\displaystyle \lim_{x \to x_0^+} (f'(x))^2=\lim_{x \to x_0^+} \frac{h''(x)}{g''(f(x))}$

h'' and g'' are different from 0, positive and finite at $\displaystyle x_0$, according to what you've written.

So :

$\displaystyle \lim_{x \to x_0^+} f'(x)=\sqrt{\frac{h''(x_0)}{g''(f(x_0))}}$

But we know that $\displaystyle f(x_0)=0$.

Thus $\displaystyle \lim_{x \to x_0^+} f'(x)=\sqrt{\frac{h''(x_0)}{g''(0)}}$
• Jul 3rd 2008, 02:02 AM
flowtsen
Quote:

Originally Posted by Moo

When you're here, why don't you multiply both sides by f'(x) ? (if I'm correct, it's not important whether f'(x)=0 or not).

$\displaystyle \lim_{x \to x_0^+} (f'(x))^2=\lim_{x \to x_0^+} \frac{h''(x)}{g''(f(x))}$

I had tried this method with a couple of examples in Mathematica, and it didn't seem to give me correct results. Are you sure that it's correct to do that manipulation?
• Jul 3rd 2008, 02:05 AM
Moo
Quote:

Originally Posted by flowtsen
I had tried this method with a couple of examples in Mathematica, and it didn't seem to give me correct results. Are you sure that it's correct to do that manipulation?

My fault, it is important whether f '(x_0)=0 or not.

If it is not, use the method above, if it is, the limit is obviously 0 (Rofl)

Do you have examples ?
• Jul 3rd 2008, 03:21 AM
flowtsen
After double-checking, this method indeed seems to work. Thanks very much, especially for the quickness of your response!