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

1. Nov 17, 2011

### DryRun

1. The problem statement, all variables and given/known data
Find limit of sin(x).sin^-1(x)/x^2 as x approaches 0

2. Relevant equations
Maybe L'Hopital's rule.

3. 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!

2. Nov 17, 2011

### ehild

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

3. Nov 17, 2011

### HallsofIvy

"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 $1/\sqrt{1- x^2}$.

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