How to Deal with Tricky Calculus Questions (Limits)

Sometimes, there are tricky questions such as the following:

$$ \lim_{x \to 3} \frac{x \; – \, 3}{\sqrt{x^2 \; – \, 9}} $$

Both the numerator and the denominator tend to \( 0 \) as \( x \) tends to \( 3 \). And \( 0/0 \) cannot be calculated. So, how can this problem be solved?

We use what is called difference of squares. Let’s calculate \( (x \; – \, 3)(x + 3) \). I can use FOIL to calculate this:

\begin{equation} \begin{split} (x \; – \, 3)(x + 3) & =x^2 + 3x \; – \, 3x \; – \, 9 \\ & = x^2 \; – \, 9 \end{split} \end{equation}

What is significant is that

$$ x^2 \; – \, 9 = (x \; – \, 3)(x + 3) $$

More generally, we have the formula:

$$ a^2 \; – \, b^2 = (a \, – \, b)(a + b) $$

this formula is known as Difference of Squares. In our case, \( a = x \) and \( b = 3 \).

Since \( x^2 \; – \, 9 = (x \; – \, 3)(x + 3) \), I can rewrite \( \frac{x \; – \, 3}{\sqrt{x^2 \; – \, 9}} \) as follows:

\begin{equation} \begin{split} \frac{x \; – \, 3}{\sqrt{x^2 \; – \, 9}} & = \frac{x \; – \, 3}{\sqrt{(x \; – \, 3)(x + 3)}} \\ & = \frac{(x \; – \, 3)/\sqrt{x \; – \, 3}}{\sqrt{x \; + \, 3}} \\ & = \frac{\sqrt{x \; – \, 3}}{\sqrt{x \; + \, 3}} \end{split} \end{equation}

Note that \( \sqrt{x \; + \, 3} \) tends to \( \sqrt{6} \) as \( x \) tends to \( 3 \) and that \( \sqrt{x \; – \, 3} \) tends to \( 0 \) as \( x \) tends to \( 3 \). So, we get:

\begin{equation} \begin{split} \lim_{x \to 3} \frac{x \; – \, 3}{\sqrt{x^2 \; – \, 9}} & = \frac{\sqrt{x \; – \, 3}}{\sqrt{x \; + \, 3}} \\ &= \frac{0}{\sqrt{6}} \\ &= 0 \end{split} \end{equation}

as our answer to this tricky question.