Math / Calculus

Question15. [-/1 Points] DETAILS MY NOTES SCALCET9M 4.4.031.
Find the limit. Use I'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. $\lim _{x \rightarrow 0} \frac{\sin ^{-1}(x)}{2 x}$ $\square$ Need Help? Read It Watch it Submit Answer

Studdy Solution
Evaluate the limit by substituting $x = 0$ into the simplified expression:
$\frac{1}{2\sqrt{1-0^2}} = \frac{1}{2\sqrt{1}} = \frac{1}{2}$
The limit is $\frac{1}{2}$ .

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