Answer
$L(x)=1$ and $Q(x)=1+\dfrac{x^2}{2!}$
Work Step by Step
We have $f(x) = \dfrac{1}{\sqrt {1-x^2}}$
and $f'(x)=\dfrac{x}{(1-x^2)^{3/2}} ; \\f''(x)=\dfrac{1}{(1-x^2)^{3/2}}+\dfrac{3x^2}{(1-x^2)^{5/2}}$
$\implies f(0)=1; f'(0)=1; f''(0)=1$
Therefore, $L(x)=f(0)+xf'(0)=1$ and $Q(x) =f(0) x +xf'(0) +\dfrac{x^2}{2!}f''(0)=1+\dfrac{x^2}{2!}$
So, $L(x)=1$ and $Q(x)=1+\dfrac{x^2}{2!}$
