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PROBLEM:
Evaluate \displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\right).
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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
\begin{align*}
\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{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\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 _{x\to 3}\left(\frac{x+3}{\sqrt{x^2-9}}\right) \\
\\
& =\frac{3+3}{\sqrt{3^2-9}} \\
\\
& =\frac{6}{0} \\
\\
& =\infty \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}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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