# Math Help - Find the Limit

1. ## Find the Limit

Find the limit as x goes to -3. (2 - Sqrt(x^2 - 5))/(x + 3)

sorry for the sloppiness if I knew the codes or where to get the latex codes from I'd write this better. Hopefully someone can write this in the proper form so its easier to read. But I know the first step is to rationalize the limit by multiplying both the numerator and denominator by (2 + sqrt(x^2 - 5)). After this I get a little confused.

2. Originally Posted by VitaX
Find the limit as x goes to -3. (2 - Sqrt(x^2 - 5))/(x + 3)

sorry for the sloppiness if I knew the codes or where to get the latex codes from I'd write this better. Hopefully someone can write this in the proper form so its easier to read. But I know the first step is to rationalize the limit by multiplying both the numerator and denominator by (2 + sqrt(x^2 - 5)). After this I get a little confused.
$\lim_{x \rightarrow -3} \frac{2-\sqrt{x^2-5}}{x+3}$

$=\lim_{x \rightarrow -3} \frac{(2-\sqrt{x^2-5})(2+\sqrt{x^2-5})}{(x+3)(2+\sqrt{x^2-5)}}$

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

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

$=\lim_{x \rightarrow -3} \frac{9-x^2}{(x+3)(2+\sqrt{x^2-5})}$

$=\lim_{x \rightarrow -3} \frac{(3-x)(3+x)}{(x+3)(2+\sqrt{x^2-5})}$

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

$=\frac{6}{4}$

$=\frac{3}{2}$

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

$=\lim_{x \rightarrow -3} \frac{(2-\sqrt{x^2-5})(2+\sqrt{x^2-5})}{(x+3)(2+\sqrt{x^2-5)}}$

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

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

$=\lim_{x \rightarrow -3} \frac{9-x^2}{(x+3)(2+\sqrt{x^2-5})}$

Oops sorry, forgot to think about the fact that the numerator factors.

4. limit in the book is 3/2 by the way. Thanks for your help i understand where i went wrong mainly in the denominator.