# [SOLVED] Limit when Hopital's rule fails

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

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

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

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

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

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!
• July 3rd 2008, 01:49 AM
Moo
Hello,

Quote:

$
\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).

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

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

So :

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

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

Thus $\lim_{x \to x_0^+} f'(x)=\sqrt{\frac{h''(x_0)}{g''(0)}}$
• July 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).

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

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?
• July 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 ?
• July 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!