Calculus 1: Find the limit as x approaches 4 of $\frac{3-\sqrt{x+5}}{x-4}$

I understand how to find limits, but for some reason I cannot figure out the algebra of this problem. I tried multiplying by the conjugate and end up with 0/0. When I check on my calculator, or apply L'Hopital's rule I get -1/6. Is there an algebra trick that I am missing on this one? $\displaystyle\frac{3-\sqrt{x+5}}{x-4}$

I have solved similar problems with the square root by multiplying by the conjugate, but it doesn't seem to work for this one.

• Note that $-x+4 = -(x-4)$ – Namaste Oct 19 '14 at 18:58

Take the limit of that as $x$ approaches $4$.
When you multiply with that conjugate, you end up with $-x+4$ in the NUM. So that "cancels" against that x-4 in the DENOM (apart from that minus) Retaking the limit gives that -1/6
$$\begin{array}{rcl}\lim_{x\to 4} \frac{3-\sqrt{x+5}}{x-4} & = & \lim_{x\to 4} \frac{(3-\sqrt{x+5})(3+\sqrt{x+5})}{(x-4)(3+\sqrt{x+5})}=\lim_{x\to 4} \frac{9-(x+5)}{(x-4)(3+\sqrt{x+5})} \\ & = & \lim_{x\to 4} \frac{4-x}{(x-4)(3+\sqrt{x+5})} =\lim_{x\to 4} \frac{-1}{3+\sqrt{x+5}}=- \frac{1}{6}.\end{array}$$