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

[sharks]

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!

[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

[HallsofIvy]

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
"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!