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I know it's pretty straight forward with L'Hopital's rule, but I was trying to solve algebraically to no avail.

$$ \lim_{x\to 2} \frac{x^2+2x - 8}{\sqrt{x^2 + 5} - (x+1)}$$

The limit is $-18$, as discerned using L'Hopital's... Can we solve algebraically?

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  • $\begingroup$ Yes!! Just rationalize the denominator. $\endgroup$ – Valerin Dec 6 '13 at 3:24
  • $\begingroup$ Ahh, I see it now. Sigh... Missed the (x-2) that comes outta the denominator to cancel. Thanks for the verification! $\endgroup$ – puppyman Dec 6 '13 at 4:16
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Yes, you can. Multiply numerator and denominator by $((x^2+5)^{1/2}+(x+1))$ and see what happens.

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My hint:

$$\lim_{x\to 2}\frac{x^2+2x-8}{\sqrt{x^2+5}-(1+x)}=\lim_{x\to 2}\frac{(x-2)(x+4)}{\frac{2(2-x)}{\sqrt{x^2+5}+1+x}}=-\lim_{x\to 2}\frac{(4+x)(\sqrt{x^2+5}+1+x)}{2}=-18$$

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