# Condition for using L'Hopital Rule

• September 10th 2011, 06:02 PM
pankaj
Condition for using L'Hopital Rule
Is there any other condition apart from the fact that limit should be of the form $\frac{0}{0},\frac{\infty}{\infty}$ so that L'Hopital Rule can be applied.

e.g. $\lim_{x\to 0}\frac{e^{\frac{1}{x}}-e^{\frac{-1}{x}}}{e^{\frac{1}{x}}+e^{\frac{-1}{x}}}$
Does not exist but when L'Hopital Rule is applied the answer comes out to be one though limit is of the type $\frac{\infty}{\infty}$.

Also $\lim_{x\to \infty}\frac{x-\sin x}{x+\sin x}=1$ but when L'Hopital Rule is applied we do not get any answer.

Can some body provide the exact condition required.Some more examples in contradiction to direct application of L'Hopital Rule will be appreciated.
• September 10th 2011, 07:34 PM
TKHunny
Re: Condition for using L'Hopital Rule
Assume f(x) / g(x) is an indeterminate form.

It is not often spoken of, but it is also required that $f^{[n]}(x)$ and $g^{[n]}(x)$ change sign only finitely many times in the neighborhood of the limit.

In your example, neither 1 - cos(x) nor 1 + cos(x) has this problem, but the next derivative certainly does.

It's also important to note that the repeated application of the rule must converge.
• September 10th 2011, 08:14 PM
Prove It
Re: Condition for using L'Hopital Rule
Quote:

Originally Posted by pankaj
Is there any other condition apart from the fact that limit should be of the form $\frac{0}{0},\frac{\infty}{\infty}$ so that L'Hopital Rule can be applied.

e.g. $\lim_{x\to 0}\frac{e^{\frac{1}{x}}-e^{\frac{-1}{x}}}{e^{\frac{1}{x}}+e^{\frac{-1}{x}}}$
Does not exist but when L'Hopital Rule is applied the answer comes out to be one though limit is of the type $\frac{\infty}{\infty}$.

Also $\lim_{x\to \infty}\frac{x-\sin x}{x+\sin x}=1$ but when L'Hopital Rule is applied we do not get any answer.

Can some body provide the exact condition required.Some more examples in contradiction to direct application of L'Hopital Rule will be appreciated.

Most indeterminate forms can use L'Hospital's Rule, but it requires a transformation to make it of the form $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$.

Indeterminate form - Wikipedia, the free encyclopedia