Question

$\lim\limits_{x \to -5} \dfrac{x^2 - 25}{x^2 + 2x - 15}$

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 - 25\) as a difference of squares: \(x^2 - 25=(x - 5)(x + 5)\).
Factor \(x^2 + 2x - 15\) by finding two numbers that multiply to \(-15\) and add to \(2\), which are \(5\) and \(-3\): \(x^2 + 2x - 15=(x + 5)(x - 3)\).
So the limit becomes:
\(\lim_{x
ightarrow -5}\frac{(x - 5)(x + 5)}{(x + 5)(x - 3)}\)

Step2: Cancel common factors

Cancel the common factor \((x + 5)\) (note \(x
eq -5\) when canceling, but we are taking the limit as \(x
ightarrow -5\), so \(x\) approaches \(-5\) but is not equal to \(-5\) at the moment of canceling):
\(\lim_{x
ightarrow -5}\frac{x - 5}{x - 3}\)

Step3: Substitute \(x = -5\)

Substitute \(x=-5\) into \(\frac{x - 5}{x - 3}\):
\(\frac{-5 - 5}{-5 - 3}=\frac{-10}{-8}=\frac{5}{4}\)

Answer:

\(\frac{5}{4}\)