Question

How to evaluate question number 4

Answer 1

There are 2 ways that spring to mind.

One is to use the definition of the derivative at a point:

$f'(c)=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

In this case, $c=0$ and $f(x)=\sqrt{5-x}$. Then

$f'(x)=-\dfrac1{2\sqrt{5-x}}\implies f'(0)=\displaystyle\lim_{x\to0}\frac{\sqrt{5-x}-\sqrt5}x=-\dfrac1{2\sqrt5}$

- - -

If you don't know about derivative yet (I would think you do, considering this is for a midterm in AP Calc, but I digress), the other way is to rely on algebraic manipulation. Multiply the numerator and denominator by the conjugate of the numerator to get

$\dfrac{\sqrt{5-x}-\sqrt5}x\cdot\dfrac{\sqrt{5-x}+\sqrt5}{\sqrt{5-x}+\sqrt5}=\dfrac{\left(\sqrt{5-x}\right)^2-\left(\sqrt5\right)^2}{x\left(\sqrt{5-x}+\sqrt5\right)}=-\dfrac x{x\left(\sqrt{5-x}+\sqrt5\right)}=-\dfrac1{\sqrt{5-x}+\sqrt5}$

This is continuous at $x=0$, so the limit is the value of the expression at $x=0$:

$\displaystyle\lim_{x\to0}\frac{\sqrt{5-x}-\sqrt5}x=-\lim_{x\to0}\frac1{\sqrt{5-x}+\sqrt5}=-\frac1{2\sqrt5}$