Today, I’ll explain how to calculate a limit that is somewhat different than the one here. Specifically, I’ll explain how to calculate this:
\begin{equation} \begin{split} \lim_{x \to 3} \frac{x^2 \; – \, 9}{\sqrt{x \; – \, 3}} = 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 \( 9 = 3^2 \). So, using difference of squares, we get
\begin{equation} \begin{split} x^2 \; – \, 9 = (x \; – \, 3) (x + 3) \end{split} \end{equation}
Hence,
\begin{equation} \begin{split} \lim_{x \to 3} \frac{x^2 \; – \, 9}{\sqrt{x \; – \, 3}} = \lim_{x \to 3} \frac{(x \; – \, 3)(x + 3)}{\sqrt{x \; – \, 3}} \end{split} \end{equation}
Notice that
\begin{equation}\begin{split} \frac{x \; – \, 3}{\sqrt{x \; – \, 3}} = \sqrt{x \; – \, 3} \end{split}\end{equation}
So,
\begin{equation} \begin{split} \lim_{x \to 3} \frac{(x \; – \, 3)(x + 3)}{\sqrt{x \; – \, 3}} = \lim_{x \to 3} \sqrt{x \; – \, 3} \, (x + 3) \end{split} \end{equation}
As \( x \) tends to \( 3 \), \( \sqrt{x \; – \, 3} \) tends to \( 0 \). So,
\begin{equation} \begin{split} \lim_{x \to 3} \frac{x^2 \; – \, 9}{\sqrt{x \; – \, 3}} & = \lim_{x \to 3} \sqrt{x \; – \, 3} \, (x + 3) \\ & = 0 \cdot 6 \\ & = 0 \end{split} \end{equation}
And there you have it!
