# Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 13|

### $\lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\right)$$\lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\right)$

SOLUTION:

A straight substitution of  $x=3$ leads to the indeterminate form   $\frac{0}{0}$   which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

$\lim\limits_{x\to \:3}\left(\frac{\sqrt{x^2-9}}{x-3}\right)=\lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\cdot \frac{\sqrt{x^2-9}}{\sqrt{x^2-9}}\right)$

$=\lim\limits_{x\to 3}\left(\frac{x^2-9}{\left(x-3\right)\sqrt{x^2-9}}\right)$

$=\lim\limits_{x\to 3}\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\sqrt{x^2-9}}\right)$

$=\lim _{x\to 3}\left(\frac{x+3}{\sqrt{x^2-9}}\right)$

$=\frac{3+3}{\sqrt{3^2-9}}$

$=\frac{6}{0}$

$=\infty$

Since the function’s limit is different from the left to its limits from the right, the limit does not exist.