# Finding $\lim_{x\to -2}{\frac{x+2}{\sqrt{-x-1}-1}}\;$ without L'Hospital

I have been trying to find $$\lim_{x\to -2}{\frac{x+2}{\sqrt{-x-1}-1}}$$ without L'Hospital's Rule, but I am stuck. I tried

• Rationalizationg the denominator
• Factoring out $\,x$

But it did not work. Finally, I used L'Hospital's Theorem and I got the answer $-2$. Is there any way to evaluate this without this concept?

Rationalizing the denominator works.

$$\lim_{x\to -2}{\frac{x+2}{\sqrt{-x-1}-1}}=\lim_{x\to -2}{\frac{x+2}{\sqrt{-x-1}-1}}.\frac{{\sqrt{-x-1}+1}}{{\sqrt{-x-1}+1}}=\lim_{x\to -2}{\frac{x+2}{-x-2}}.{(\sqrt{-x-1}+1)}=\lim_{x\to -2}-{(\sqrt{-x-1}+1)}=-2$$

• Sorry, posted just after the other answer. – GoodDeeds Feb 5 '16 at 11:25

If you rationalize the denominator becomes $-(x+2)$ and after the cancellation all that remains is $$-(\sqrt{-x-1}+1)$$

Now if you take the limit you get $-2$.

Here is an easier solution by substitution: use $u = \sqrt{-x-1} \Rightarrow x = -(u^2+1)$ the original formula:

$$\lim_{x\to -2}{\frac{x+2}{\sqrt{-x-1}-1}}$$

$$\lim_{u\to 1}{\frac{-(u^2+1)+2}{u-1}}$$

$$\lim_{u\to 1}{\frac{-(u-1)(u+1)}{u-1}}$$

since $\lim_{y\to 0}{\frac{y}{y}} = 1$ then the $$\lim_{u\to 1}{\frac{-(u-1)(u+1)}{u-1}} = \lim_{u\to 1}{\frac{-(u+1)}{1} = -2}$$

$$\frac{x+2}{\sqrt{-x-1}-1} = \frac{x+2}{\sqrt{-x-1}-1}\bigg(\frac{\sqrt{-x-1}+1}{\sqrt{-x-1}+1}\bigg)$$

$$= \frac{(x+2)(\sqrt{-x-1}+1)}{-(x+2)} = -1-\sqrt{-x-1}$$

$$\lim_{x\rightarrow -2}\bigg(-1-\sqrt{-x-1}\bigg) = -2$$

HINT:

Set $\sqrt{-x-1}=y\implies y^2=-x-1\iff y^2\to1\implies y\ne1$