# L'Hopital Rule

1. Sep 28, 2010

### Kreizhn

1. The problem statement, all variables and given/known data
Find
$$\lim_{x\to0} \frac{\arcsin(x)-x}{x^3}$$

3. The attempt at a solution
This is obviously an indeterminate form, so we apply L'hopital's rule to get

$$\lim_{x\to0} \frac{\frac1{\sqrt{1-x^2}} - 1 }{ 3x^3}$$
which is again an indeterminate form so we apply it again to get
$$\lim_{x\to0} \frac{(1-x^2)^{-\frac32}}6 [/itex] from which the solution is obviously $\frac16$. However, this is my question. After the first application of L'Hopital, we could have simplified [tex] \lim_{x\to0} \frac{\frac1{\sqrt{1-x^2}} - 1 }{ 3x^3} = \lim_{x\to0} \frac{1-\sqrt{1-x^2}}{3x^2\sqrt{1-x^2}}$$
This is no longer an indeterminate form and would suggest that the limit does not exist. Is there any justification for why this can't be done? Possibly, do we know that either: 1) This simplification is not permitted after applying L'Hopital or 2) We know the limit exists and is finite and so are forced to apply L'Hopital yet again?

2. Sep 28, 2010

### Dick

Why do you think lim x->0 of (1-sqrt(1-x^2))/(3*x^2*sqrt(1-x^2)) isn't indeterminant? It looks like 0/0 to me.

3. Sep 28, 2010

### Kreizhn

Haha, yes. The ability to subtract has apparently escaped me. Thanks