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

  • Thread starter DryRun
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  • #1
DryRun
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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!
 

Answers and Replies

  • #2
ehild
Homework Helper
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Write out the function as

[tex]\frac{sin(x)}{x}\frac{sin^{-1}(x)}{x}[/tex].

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

ehild
 
  • #3
HallsofIvy
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Homework Helper
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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 [tex]sin^{-1}(x)[/tex] means [tex]1/sin(x)[/tex]. That's clearly NOT the case here since, if it were, "[itex]sin(x)sin^{-1}(x)[/itex]" would just be 1 and [itex]sin(x)sin^{-1}(x)/x^2[/itex] would be just [itex]1/x^2[/itex] which does go to infinity.

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

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

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