Originally Posted by

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

e.g.$\displaystyle \lim_{x\to 0}\frac{e^{\frac{1}{x}}-e^{\frac{-1}{x}}}{e^{\frac{1}{x}}+e^{\frac{-1}{x}}}$

Does not exist but when L'Hopital Rule is applied the answer comes out to be one though limit is of the type $\displaystyle \frac{\infty}{\infty}$.

Also $\displaystyle \lim_{x\to \infty}\frac{x-\sin x}{x+\sin x}=1$ 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.