There are 2 ways that spring to mind.
One is to use the definition of the derivative at a point:
[tex]f'(c)=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}[/tex]
In this case, [tex]c=0[/tex] and [tex]f(x)=\sqrt{5-x}[/tex]. Then
[tex]f'(x)=-\dfrac1{2\sqrt{5-x}}\implies f'(0)=\displaystyle\lim_{x\to0}\frac{\sqrt{5-x}-\sqrt5}x=-\dfrac1{2\sqrt5}[/tex]
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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
[tex]\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}[/tex]
This is continuous at [tex]x=0[/tex], so the limit is the value of the expression at [tex]x=0[/tex]:
[tex]\displaystyle\lim_{x\to0}\frac{\sqrt{5-x}-\sqrt5}x=-\lim_{x\to0}\frac1{\sqrt{5-x}+\sqrt5}=-\frac1{2\sqrt5}[/tex]
