# Find limit of sin(x).sin^-1(x)/x^2 as x approaches 0

Gold Member

## Homework Statement

Find limit of sin(x).sin^-1(x)/x^2 as x approaches 0

## Homework Equations

Maybe L'Hopital's rule.

## The Attempt at a Solution

First, i put in zero, and got the indeterminate form 0/0
So, i used L'Hopital's rule and got this:

[cosx.sin^-1(x) + sinx.-sin^-2(x).cosx]/2x

And again when putting zero, i get [1/0 + 0]/0 = infinity/0
So i tried to differentiate again, but it's become so big, i don't know what's what in my copybook. Help!

ehild
Homework Helper
Write out the function as

$$\frac{sin(x)}{x}\frac{sin^{-1}(x)}{x}$$.

You know the limit of the first fraction. Apply L'Hospital to the second one.

ehild

HallsofIvy
Homework Helper

## Homework Statement

Find limit of sin(x).sin^-1(x)/x^2 as x approaches 0

## Homework Equations

Maybe L'Hopital's rule.

## The Attempt at a Solution

First, i put in zero, and got the indeterminate form 0/0
So, i used L'Hopital's rule and got this:

[cosx.sin^-1(x) + sinx.-sin^-2(x).cosx]/2x

And again when putting zero, i get [1/0 + 0]/0 = infinity/0
"infinity/0" is NOT an indeterminate form! That would tell you the limit is infinity.

However, you seem to be thinking that $$sin^{-1}(x)$$ means $$1/sin(x)$$. That's clearly NOT the case here since, if it were, "$sin(x)sin^{-1}(x)$" would just be 1 and $sin(x)sin^{-1}(x)/x^2$ would be just $1/x^2$ which does go to infinity.

However, here, $sin^{-1}(x)$ is the arcsine, the inverse function to sine. The derivative of arcsin(x) is $1/\sqrt{1- x^2}$.

So i tried to differentiate again, but it's become so big, i don't know what's what in my copybook. Help!