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

Evaluate

\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. 

 

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