TITLE: What am I doing wrong in calculating the following limit?
QUESTION [7 upvotes]: $$\lim_{x\to-2} \frac{x+2}{\sqrt{6+x}-2}=\lim_{x\to-2} \frac{1+2/x}{\sqrt{(6/x^2)+(1/x)}-2/x^2}$$ Dividing numerator and denominator by $x \neq0$
$$\frac{1+2/-2}{\sqrt{(6/4)+(1/-2)}-2/4}=\frac{0}{1/2}=0$$ but the limit is $4$ according to Wolfram Alpha?

REPLY [3 votes]: You can apply L'Hospital rule, i.e differentiating numerator and denominator functions with respect to $x$,
$$2\sqrt{x + 6} = 2 \sqrt{-2 + 6} = 4$$