## University Calculus: Early Transcendentals (3rd Edition)

$L(x)=x$
$f(x)=\sin^{-1}x$, $L(x)=f(a)+f'(a)(x-a)$ $f'(x)=\frac{1}{\sqrt{(1-x^2)}}$ $x_{0}=0$ $f(0)={\sin^{-1}0}=0$ $f'(0)= \frac{1}{1} = 1$ then $L(x)=0+1(x-0)=x$ $L(x)=x$ Thus, the final answer is: $L(x)=x$