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

Differential and Integral Calculus by Feliciano and Uy Solution Manual by


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



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)




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