## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0}\frac{\sin^{-1}x}{x}=1$$
$$A=\lim_{x\to0}\frac{\sin^{-1}x}{x}$$ Because $\lim_{x\to0}(\sin^{-1}x)=\sin^{-1}0=0$ and $\lim_{x\to0}x=0,$ an indeterminate form of $\frac{0}{0}$ is here, so L'Hospital's Rule can be applied. $$A=\lim_{x\to0}\frac{(\sin^{-1}x)'}{x'}$$ We have $(\sin^{-1}x)'=\frac{1}{\sqrt{1-x^2}}$, hence $$A=\lim_{x\to0}\frac{\frac{1}{\sqrt{1-x^2}}}{1}$$ $$A=\lim_{x\to0}\frac{1}{\sqrt{1-x^2}}$$ $$A=\frac{1}{\sqrt{1-0^2}}$$ $$A=\frac{1}{\sqrt1}=1$$