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

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    $\begingroup$ Why do you have $2/x^2$ in denominator? $\endgroup$ – Abstraction May 23 '16 at 9:16
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    $\begingroup$ Standard trick for this kind of limit: multiply numerator and denominator by $\sqrt{6+x}+2$. $\endgroup$ – TastyRomeo May 23 '16 at 9:19
  • $\begingroup$ Spotted my mistake thank you. $\endgroup$ – Samuel May 23 '16 at 9:23
  • $\begingroup$ Try math.stackexchange.com/questions/1789907/… $\endgroup$ – lab bhattacharjee May 23 '16 at 9:28

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

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    $\begingroup$ Its been a while since I did limits so forgive me if this is a silly question but when taking the square root in the final step is there a reason we take the positive root and discard the negative root? $\endgroup$ – Chris May 23 '16 at 11:44
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    $\begingroup$ That's a matter of definition, $x^2 = a$ has two solutions (for $a>0$) but the notation $\sqrt{a}$ is reserved for the positive one; so $\sqrt{4} = 2$. $\endgroup$ – StackTD May 23 '16 at 11:56
  • $\begingroup$ Ah, cool. That makes sense. Thanks. :) $\endgroup$ – Chris May 23 '16 at 12:35
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    $\begingroup$ In fact, if you plug $x=-2$ into the denominator of your last expression in the first formula, you get $\sqrt{\tfrac6{x^2}+\tfrac1x}-\tfrac2x=1+1=2$. So this would lead to limit equal to zero. The correct version would be, in my opinion, $-\sqrt{\tfrac6{x^2}+\tfrac1x}-\tfrac2x$ in the denominator. Notice that this leads to the indeterminate form $\tfrac00$, as expected. (We are using that $\sqrt{x^2}=|x|=-x$ for $x<0$. Since we are interested in the values close to $-2$, it suffices to work with negative $x$.) $\endgroup$ – Martin Sleziak May 23 '16 at 13:33
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    $\begingroup$ @MrReality Maybe this is what you're after: $\frac{\sqrt{6+x}}{x}= \frac{\sqrt{6+x}}{-\sqrt{x^2}}= -\frac{\sqrt{6+x}}{\sqrt{x^2}}= -\sqrt{\frac{6+x}{x^2}}$. In any case, we're leaving too many irrelevant comments here and additionally each of the adds ping to the answerer. So perhaps it would be better to continue in chat if further comments are needed. $\endgroup$ – Martin Sleziak Oct 26 '17 at 19:01

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

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    $\begingroup$ ;| quite fast $\endgroup$ – user5954246 May 23 '16 at 9:21

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

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


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