Answer
$L(x)=-0.1+2.5x$
Work Step by Step
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, $L(x)=(-0.1)+(x)(2.5) =-0.1+2.5x$
