"infinity/0" is NOT an indeterminate form! That would tell you the limit is infinity.sharks said: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!
The limit of sin(x).sin^-1(x)/x^2 as x approaches 0 is equal to 1/2.
To find the limit of sin(x).sin^-1(x)/x^2 as x approaches 0, we can use the L'Hopital's Rule or the trigonometric identity sin(x)/x = 1 as x approaches 0.
This limit is equal to 1/2 because as x approaches 0, the value of sin(x)/x approaches 1. Therefore, the limit becomes sin^-1(x)/x, which can be rewritten as (sin^-1(x)/x) * (1/x). As x approaches 0, the first term becomes 1 and the second term becomes infinity, resulting in a limit of 1/2.
Yes, this limit can also be evaluated using the Taylor Series expansion for sin^-1(x) and x^2 as x approaches 0. In this method, we can also get a limit of 1/2.
Finding the limit of this expression helps us understand the behavior of trigonometric functions and how they approach a certain value as the input approaches a particular point. It also has applications in calculus and other branches of mathematics.