What am I doing wrong in calculating the following limit?

Question

$$\lim_{x\to-2} \frac{x+2}{\sqrt{6+x}-2}=\lim_{x\to-2} \frac{1+2/x}{\sqrt{(6/x^2)+(1/x)}-2/x^2}$$ 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?

Answer 1

You say you divide by $x$, but that's not what you do in the denominator; it would be:

$$\lim_{x\to-2} \frac{x+2}{\sqrt{6+x}-2} =\lim_{x\to-2} \frac{1+\tfrac{2}{x}}{\tfrac{\sqrt{6+x}}{x}-\tfrac{2}{x}} =\lim_{x\to-2} \frac{1+\tfrac{2}{x}}{-\sqrt{\tfrac{6}{x^2}+\tfrac{1}{x}}-\tfrac{2}{x}} $$

A better approach: $$\begin{array}{rl} \displaystyle \lim_{x\to-2} \frac{x+2}{\sqrt{6+x}-2} & \displaystyle = \lim_{x\to-2} \frac{\left(x+2\right)\color{blue}{\left(\sqrt{6+x}+2\right)}}{\left(\sqrt{6+x}-2\right)\color{blue}{\left(\sqrt{6+x}+2\right)}} \\[7pt] & \displaystyle = \lim_{x\to-2} \frac{\left(x+2\right)\left(\sqrt{6+x}+2\right)}{x+2} \\[7pt] & \displaystyle = \lim_{x\to-2} \left(\sqrt{6+x}+2\right) \\ & = 4 \end{array}$$

Answer 2

One may set $u=x+2$, then, as $x \to -2$, we have $u \to 0$, giving $$ \frac{x+2}{\sqrt{6+x}-2}=\frac{u}{\sqrt{u+4}-2}\times\frac{\sqrt{u+4}+2}{\sqrt{u+4}+2}=\sqrt{u+4}+2 \to 4. $$

Answer 3

You can apply L'Hospital rule, i.e differentiating numerator and denominator functions with respect to $x$,

$$2\sqrt{x + 6} = 2 \sqrt{-2 + 6} = 4$$