# Can this limit be solved algebraically?

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?

• Yes!! Just rationalize the denominator. – Valerin Dec 6 '13 at 3:24
• Ahh, I see it now. Sigh... Missed the (x-2) that comes outta the denominator to cancel. Thanks for the verification! – puppyman Dec 6 '13 at 4:16

Yes, you can. Multiply numerator and denominator by $((x^2+5)^{1/2}+(x+1))$ and see what happens.

My hint:

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