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# Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.$\displaystyle \lim_{x\to 0} \frac{\sin^{-1} x}{x}$

## This limit has the form$\frac{0}{0} \cdot \lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x} \stackrel{\Perp}{=} \lim _{x \rightarrow 0} \frac{1 / \sqrt{1-x^{2}}}{1}=\lim _{x \rightarrow 0} \frac{1}{\sqrt{1-x^{2}}}=\frac{1}{1}=1$

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Okay. So here what we have is we're wanting to use looked house rule to find the limit. So we have the inverse sine of X over X. So we see um the arc sine of X over X. When we plug in zero, we get 0/0. So that's ended giving us um the indeterminate form. So instead what we have to do is take the derivative of the numerator and the denominator separately. So we're going to get one over the square root of one minus X squared Divided by one. So that's just going to go away. Now if we plug in zero, what we end up getting as a result is one Over the square root of 1 0, which is just one over the square to one, which is just 1/1, which gives us one as the final answer for this particular limit. And that means that even though it looks complicated at first using inverse sine, we can simplify things with local styles rule to find the women

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