Using Difference of Squares for Limits

Today, I’ll explain how to calculate another limit that is somewhat different than the one here. Specifically, I’ll explain how to calculate this:

\begin{equation} \begin{split} \lim_{x \to 6} \frac{x \; – \, 6}{\sqrt{x^2 \; – \, 36}} = 0 \end{split} \end{equation}

Notice that both the numerator and the denominator in the above limit both tend to zero. To resolve this issue, notice that \( 36 = 6^2 \). So, using difference of squares, we get

\begin{equation} \begin{split} x^2 \; – \, 36 = (x \; – \, 6) (x + 6) \end{split} \end{equation}

Hence,

\begin{equation} \begin{split} \lim_{x \to 6} \frac{x \; – \, 6}{\sqrt{x^2 \; – \, 36}} = \lim_{x \to 6} \frac{x \, – \, 6}{\sqrt{(x \, – \, 6)(x + 6)}} \end{split} \end{equation}

Notice that

\begin{equation}\begin{split} \frac{x \; – \, 6}{\sqrt{x \, – \, 6}} = \sqrt{x \; – \, 6} \end{split}\end{equation}

So,

\begin{equation} \begin{split} \lim_{x \to 6} \frac{x \; – \, 6}{\sqrt{(x \; – \, 6)(x + 6) }} = \lim_{x \to 6} \frac{\sqrt{x \; – \, 6}}{x + 6} \end{split} \end{equation}

As \( x \) tends to \( 6 \), \( \sqrt{x \; – \, 6} \) tends to \( 0 \). So,

\begin{equation} \begin{split} \lim_{x \to 6} \frac{x \; – \, 6}{\sqrt{x^2 \; – \, 36}} & = \lim_{x \to 6} \frac{\sqrt{x \; – \, 6}}{x + 6} \\ & = \frac{0}{12} \\ & = 0 \end{split} \end{equation}

And there you have it!