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
\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&#039;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&#039;t be done? Possibly, do we know that either: 1) This simplification is not permitted after applying L&#039;Hopital or 2) We know the limit exists and is finite and so are forced to apply L&#039;Hopital yet again?

 

[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