Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 13

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PROBLEM:

Evaluate lim⁡x→3(x2−9x−3)\displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\right).

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SOLUTION:

A straight substitution of x=3x=3 leads to the indeterminate form 00\frac{0}{0} which is meaningless.

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

lim⁡x→ 3(x2−9x−3)=lim⁡x→3(x2−9x−3⋅x2−9x2−9)=lim⁡x→3((x+3)(x−3)(x−3)x2−9)=lim⁡x→3(x2−9(x−3)x2−9)=lim⁡x→3(x+3x2−9)=3+332−9=60=∞  (Answer)\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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