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

$L(x)=-0.1+2.5x$
The linear approximation of $f(x)$ at $x=0$ is given as: $L(x)=f(0)+x f'(0)$ We are given that $f(x)=\dfrac{2}{1-x}+\sqrt {1-x}-3.1$ and $f(0)=-0.1$ Also, $f'(x)=\dfrac{1}{(1-x)^2}+\dfrac{1}{2 \sqrt {1+x}}$ and $f'(0)=2.5$ Hence, $L(x)=(-0.1)+(x)(2.5) =-0.1+2.5x$