Solving $\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$ without L'Hopital. I am trying to solve the limit
$$\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$$
Without using L'Hopital.
Evaluating yields $\frac{0}{0}$. When I am presented with roots, I usually do this:
$$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}} \cdot \frac{3+\sqrt{5+x}}{3+\sqrt{5+x}}$$
And end up with
$$\frac{9 - (5+x)}{(1-\sqrt{5-x}) \cdot (3+\sqrt{5+x})}$$
Then
$$\frac{9 - (5+x)}{3+\sqrt{5+x}-3\sqrt{5-x}-(\sqrt{5-x}\cdot\sqrt{5+x})}$$
And that's
$$\frac{0}{3+\sqrt{9}-3\sqrt{1}-(\sqrt{1}\cdot\sqrt{9})}= \frac{0}{3+3-3-3-3} = \frac{0}{-3} = -\frac{0}{3}$$
Edit: Ok, the arithmetic was wrong. That's $\frac{0}{0}$ again.
But the correct answer is
$$-\frac{1}{3}$$
But I don't see where does that $1$ come from.
 A: I would rewrite the quotient as the quotient of two variation rates. Set $x=4+h\enspace(h\to 0)$:
\begin{align*}
\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}&=\frac{\sqrt{9}-\sqrt{9+h}}{\sqrt{1}-\sqrt{1-h}}\\
&=\frac{\sqrt{9+h}-\sqrt{9}}h\cdot\frac h{\sqrt{1-h}-\sqrt{1}}\to\bigl(\sqrt{x}\bigr)'_{x=9}\cdot-\frac1{\bigl(\sqrt{x}\bigr)'_{x=1}}\\
&=-\frac1{2\sqrt 9}\cdot2\sqrt{1}=-\frac13.
\end{align*}
A: For every $x\in [-5,5]\setminus\{4\}$ we have:
\begin{eqnarray}
\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}&=&\frac{(3-\sqrt{5+x})\color{green}{(3+\sqrt{5+x})}\color{blue}{(1+\sqrt{5-x})}}{(1-\sqrt{5-x})\color{blue}{(1+\sqrt{5-x})}\color{green}{(3+\sqrt{5+x})}}=\frac{[9-(5+x)](1+\sqrt{5-x})}{[1-(5-x)](3+\sqrt{5+x})}\\
&=&\frac{-\color{red}{(x-4)}(1+\sqrt{5-x})}{\color{red}{(x-4)}(3+\sqrt{5+x})}=-\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}},
\end{eqnarray}
it follows that
$$
\lim_{x\to4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\lim_{x\to4}\left(-\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}\right)=-\frac{1+1}{3+3}=-\frac13.
$$
A: $$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\frac{\frac{x-4}{-(3-\sqrt{5+x})}}{\frac{x-4}{1+\sqrt{5-x}}}=\frac{1+\sqrt{5-x}}{-(3+\sqrt{5+x})}$$
