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.
 A: 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 
A: $$\begin{array}{rcl}\lim_{x\to 4} \frac{3-\sqrt{x+5}}{x-4} & = & \lim_{x\to 4} \frac{(3-\sqrt{x+5})(3+\sqrt{x+5})}{(x-4)(3+\sqrt{x+5})}=\lim_{x\to 4} \frac{9-(x+5)}{(x-4)(3+\sqrt{x+5})} \\ & = & \lim_{x\to 4} \frac{4-x}{(x-4)(3+\sqrt{x+5})} =\lim_{x\to 4} \frac{-1}{3+\sqrt{x+5}}=- \frac{1}{6}.\end{array}$$
A: \begin{gathered}
  \frac{{3 - \sqrt {x + 5} }}{{x - 4}} = \frac{{\left( {3 - \sqrt {x + 5} } \right)\left( {3 + \sqrt {x + 5} } \right)}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\ 
   = \frac{{{3^2} - {{\sqrt {x + 5} }^2}}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\ 
   = \frac{{9 - (x + 5)}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\ 
   = \frac{{ - (x - 4)}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\ 
   = \frac{{ - 1}}{{\left( {3 + \sqrt {x + 5} } \right)}} \\ 
\end{gathered}
Take the limit of that as $x$ approaches $4$.
