# l'Hospital rule

The condition for l'Hospital rule is that after substituting variables with values the limit has to be in the form

$$\frac00\quad\text{or}\quad \frac\infty\infty$$

1. $\displaystyle\lim_{x\to 0} x^2 \sin x$

2. $\displaystyle\lim_{x\to 1}\frac{x^m-1}{x^n-1}$ where $m,n\in\mathbb{N}$

• First limit is straight forward. $\displaystyle\lim_{x\to 0} x^2 \sin x$=0 but what do you think about $\displaystyle\lim_{x\to 0} x^2 cosec x$ ? – Mathematics Dec 20 '15 at 12:21

One often forgets an essential condition to apply L'Hospital's rule: the denominator must not be $0$ in some neighbourhood of the point at which we compute the limit, except at this point.
• Usually, there is no problem: most functions have isolated roots. You may have difficulties with functions like $\; x\sin \dfrac1x$ in the neighbourhood of $0$. There is no general strategy, as far as I know. Let me add, most of the tima L'Hospital's rule should be avoided: it is logically equivalent to using Taylor's formula at order $1$, which doesn't have this drawback. – Bernard Dec 20 '15 at 13:14
For the first limit, the limit is simply $0$, there is no fraction. For, the second, yes it is of the form $0/0$ and you can apply L'Hopital's rule since both terms - numerator and denominator - are polynomials and hence differentiable. It will give you $$\lim_{x\to1}\frac{x^m-1}{x^n-1}\overset{\frac00}=\lim_{x\to1}\frac{(x^m-1)'}{(x^n-1)'}=\lim_{x\to1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m\cdot1}{n\cdot1}=\frac{m}{n}$$ for $m,n \in \mathbb N_{>0}$.
• For the second one, I think that the limit is just $\frac mn$ – Claude Leibovici Dec 20 '15 at 12:56
• @ClaudeLeibovici You are correct. I was thinking about $x\to \infty$. Let me edit my answer. – Jimmy R. Dec 20 '15 at 12:57