Solving $\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$ without L'Hopital.

Question

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.

Answer 1

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 $$ \lim_{x\to4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\lim_{x\to4}\left(-\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}\right)=-\frac{1+1}{3+3}=-\frac13. $$

Answer 2

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*}

Answer 3

$$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\frac{\frac{x-4}{-(3-\sqrt{5+x})}}{\frac{x-4}{1+\sqrt{5-x}}}=\frac{1+\sqrt{5-x}}{-(3+\sqrt{5+x})}$$