To find the limit of [tex]\(\frac{x}{\sin^{-1} x}\)[/tex] as [tex]\(x\)[/tex] approaches 0, we can follow these steps:
1. Understand the Expression:
We need to evaluate [tex]\(\lim_{x \to 0} \frac{x}{\sin^{-1} x}\)[/tex].
2. Behavior at the Limit Point:
As [tex]\(x\)[/tex] approaches 0, both [tex]\(x\)[/tex] and [tex]\(\sin^{-1} x\)[/tex] approach 0. This suggests we might use L'Hôpital's Rule, which is applicable for limits of the form [tex]\(\frac{0}{0}\)[/tex].
3. Apply L'Hôpital's Rule:
L'Hôpital's Rule states that if [tex]\(\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0}\)[/tex] or [tex]\(\frac{\pm \infty}{\pm \infty}\)[/tex], then
[tex]\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)},
\][/tex]
provided the limit on the right side exists.
Here, let [tex]\(f(x) = x\)[/tex] and [tex]\(g(x) = \sin^{-1} x\)[/tex]. Both functions are differentiable around [tex]\(x = 0\)[/tex].
4. Differentiate the Numerator and Denominator:
- The derivative of [tex]\(f(x) = x\)[/tex] is [tex]\(f'(x) = 1\)[/tex].
- The derivative of [tex]\(g(x) = \sin^{-1} x\)[/tex] is [tex]\(g'(x) = \frac{1}{\sqrt{1 - x^2}}\)[/tex].
5. Apply L'Hôpital's Rule:
Therefore,
[tex]\[
\lim_{x \to 0} \frac{x}{\sin^{-1} x} = \lim_{x \to 0} \frac{1}{\frac{1}{\sqrt{1 - x^2}}}.
\][/tex]
6. Simplify the Expression:
Simplify the fraction:
[tex]\[
\lim_{x \to 0} \frac{1}{\frac{1}{\sqrt{1 - x^2}}} = \lim_{x \to 0} \sqrt{1 - x^2}.
\][/tex]
7. Evaluate the Limit:
As [tex]\(x\)[/tex] approaches 0, the expression [tex]\(\sqrt{1 - x^2}\)[/tex] approaches [tex]\(\sqrt{1 - 0} = 1\)[/tex].
Therefore,
[tex]\[
\lim_{x \rightarrow 0} \frac{x}{\sin^{-1} x} = 1.
\][/tex]
So, the limit is [tex]\(\boxed{1}\)[/tex].
