# Finding the linear approximation of $\frac{1}{\sqrt{2-x}}$ at $x=0$

The linear approximation at $x=0$ to $\dfrac{1}{\sqrt{2-x}}$ is $A+Bx$, where $A$ is: _____, and where $B$ is: ______.

I don't understand what this question is asking and how to solve it. I know how to use linear approximations to estimate a single number, but what am I supposed to do here? I tried taking the derivative and having $A= \dfrac{1}{2(2-x)^{3/2}},$ but that is wrong.

Thanks.

• $A=f(0)$ and $B=f^{\prime}(0)$. Oct 6 '14 at 20:48
– Karl
Oct 6 '14 at 21:02
• @user84413 would $f\;'(0)=1/4^{3/2}$? Oct 6 '14 at 21:32
• I think it should be $\frac{1}{2(2-x)^{3/2}}$ evaluated at $x=0$. Oct 6 '14 at 22:31

Recall that the linear approximation of $f(x)$ at $x=a$ is given by $$\ell(x)=f'(a)(x-a)+f(a).$$
Here, $f(x)={1\over \sqrt{2-x}}$ and $a=0$, so \begin{align} f(x)&=(1-2x)^{-1/2}\implies f(0)=1\\ f'(x)&=-{1\over 2}(1-2x)^{-3/2}(-2)={-1\over (1-2x)^{3/2}}\implies f'(0)=-1,\\ \end{align} thus $$\ell(x)=-1(x-0)+1=1-x.$$ Therefore, $A=1$ and $B=-1$.
$\textbf{Hint}$
$$\left(a + x\right)^{-1/2} = a^{-1/2}\left(1+\dfrac{x}{a}\right)^{-1/2}$$ and use expansion since $\dfrac{x}{a}<<1$ in the limit $x \rightarrow 0$