Question

Find the limit:

Answer 1

\[\lim_{x \rightarrow 0}~\frac{\sin^{-1}x}{x}\]

Answer 2

I would use L'Hopital's Rule if I were you.

Answer 3

Do you know how to apply L'Hopital's Rule?

Answer 4

Yea, just had to remember what \(\frac{d}{dx}sin^{-1}x\) was..

So with L'Hopital I get \(\LARGE \lim_{x \rightarrow 0} \frac{1}{\sqrt{1-x^2}}\)?

Answer 5

Yes, now all you have to do is plug in 0.

Answer 6

Alright, thank you!

Answer 7

Glad I helped!