Sunday, September 17, 2017

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.



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


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