calculus - Finding $lim_{xto -2}{frac{x+2}{sqrt{-x-1}-1}};$ 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?

Answer

Rationalizing the denominator works.

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