How to Use Difference of Squares

Today, I’ll explain how to calculate the following limit:

\begin{equation} \begin{split} \lim_{x \to 5} \frac{x^2 \; – \, 25}{x \; – \, 5} = 10 \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 \( 25 = 5^2 \). So, using difference of squares, we get

\begin{equation} \begin{split} x^2 \; – \, 25 = (x \; – \, 5) (x + 5) \end{split} \end{equation}

Hence,

\begin{equation} \begin{split} \lim_{x \to 5} \frac{x^2 \; – \, 25}{x \; – \, 5} = \lim_{x \to 5} \frac{(x \; – \, 5)(x + 5)}{x \; – \, 5} \end{split} \end{equation}

We can cancel out \( x \, – \, 5 \) in both the numerator and the denominator to get

\begin{equation} \begin{split} \lim_{x \to 5} \frac{(x \; – \, 5)(x + 5)}{x \; – \, 5} = \lim_{x \to 5} x + 5 \end{split} \end{equation}

So,

\begin{equation} \begin{split} \lim_{x \to 5} \frac{x^2 \; – \, 25}{x \; – \, 5} = \lim_{x \to 5} x + 5 = 10 \end{split} \end{equation}

And there you have it!