Math Series #14

For $x \neq 5,$ $\frac{x^2-25}{x-5} = x+5$ .

This is a linear function with a removable discontinuity at $x=5$ .

Thus $\lim_{x \rightarrow 5} \frac{x^2-25}{x-5} = \lim_{x \rightarrow 5} (x+5) = 10.$

We can't use direct substitution as it gives $\dfrac{0}{0}$

$\lim_{x \rightarrow 5} \dfrac{x^{2} - 25}{x - 5} = \lim_{x \rightarrow 5} \dfrac{(x+5)(x-5)}{x - 5} = \lim_{x \rightarrow 5} x + 5 = 5 + 5$ $\therefore \boxed{10}$

If we insert $x = 5$ into the equation, $\frac {x^{2}-25}{x-5} = \frac {0}{0}$ , which is undefined. We can factor $x^{2}-25$ into $(x+5)(x-5)$ . Since $\frac {(x+5)(x-5)}{x-5} = x+5$ , $\lim_{x\to5} x+5 = 5 + 5 = \boxed{10}$