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I need to evaluate the following limit:

$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$

I have multiplied both sides by the conjugate $\sqrt{x^2+5}+3$ but am getting $x^2-4$ as the denominator. Is this the correct way to go about it?

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  • $\begingroup$ Have you considered l'Hopital's rule? $\endgroup$ – Bonnaduck Oct 13 '13 at 15:37
  • $\begingroup$ 10-11 minutes. $ $ $\endgroup$ – Did Oct 13 '13 at 15:55
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Yes indeed, that's the way to go about it. Now, we have $$\begin{align} \lim_{x \to 2}\frac{x-2}{\sqrt{x^2+5}-3} & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{x^2 - 4} \\ \\ & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{(x - 2)(x+2)} \\ \\ & \overset{x\neq 2}{=} \lim_{x \to 2}\dfrac {\sqrt{x^2 + 5} + 3}{x+2} \\ \\ & = \dfrac 64 = \frac 32\end{align}$$

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From what you have got, you can simply cancel the factor $(x-2)$ from both numerator and denominator. After that, you can substitute $x=2$ into the expression and obtain the limit.

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You are on the right track.

$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}=\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3} \frac{\sqrt{x^2+5}+3}{\sqrt{x^2+5}+3}=\lim_{x\to 2} \frac{(x-2)(\sqrt{x^2+5}+3)}{(x-2)(x+2)}$ simplifying $$\lim_{x\to 2} \frac{(\sqrt{x^2+5}+3)}{(x+2)}$$ and evaluating at $2$ we get

$$\lim_{x\to 2} \frac{(\sqrt{x^2+5}+3)}{(x+2)}= \frac{(\sqrt{4+5}+3)}{(2+2)}=\frac{6}{4}=\frac{3}{2}$$

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In $\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$ let $x = y+2$.

This becomes

$\begin{align} \lim_{y\to 0} \frac{y}{\sqrt{(y+2)^2+5}-3} &=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+4+5}-3}\\ &=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+9}-3}\\ &=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+9}-3}\frac{\sqrt{y^2+4y+9}+3}{\sqrt{y^2+4y+9}+3}\\ &=\lim_{y\to 0} \frac{y(\sqrt{y^2+4y+9}+3)}{(\sqrt{y^2+4y+9}-3)(\sqrt{y^2+4y+9}+3)}\\ &=\lim_{y\to 0} \frac{y(\sqrt{y^2+4y+9}+3)}{(y^2+4y+9)-9}\\ &=\lim_{y\to 0} \frac{y(\sqrt{y^2+4y+9}+3)}{y^2+4y}\\ &=\lim_{y\to 0} \frac{\sqrt{y^2+4y+9}+3}{y+4}\\ &=\frac{\sqrt{9}+3}{4}\\ &=\frac{6}{4}\\ &=\frac{3}{2}\\ \end{align} $

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