Does Simplifying After Applying L'Hopital Rule Affect the Limit?

[Kreizhn]

Homework Statement

Find
$$\lim_{x\to0} \frac{\arcsin(x)-x}{x^3}$$

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
[tex]\lim_{x\to0} \frac{(1-x^2)^{-\frac32}}6 [/itex]<br /> from which the solution is obviously $\frac16$.<br /> <br /> However, this is my question. After the first application of L'Hopital, we could have simplified<br /> $$\lim_{x\to0} \frac{\frac1{\sqrt{1-x^2}} - 1 }{ 3x^3} = \lim_{x\to0} \frac{1-\sqrt{1-x^2}}{3x^2\sqrt{1-x^2}}$$<br /> 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?[/tex]

 

[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.

 

[Kreizhn]

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