3. $lim_{x\rightarrow - 5}\frac{x^{2}+2x - 15}{x^{2}-25}=$

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}+2x - 15=(x + 5)(x - 3)$ and the denominator $x^{2}-25=(x + 5)(x - 5)$. So the limit becomes $\lim_{x\rightarrow - 5}\frac{(x + 5)(x - 3)}{(x + 5)(x - 5)}$.

Step2: Cancel out the common factor

Since $x\neq - 5$ when taking the limit (we are approaching - 5, not evaluating at - 5), we can cancel out the $(x + 5)$ terms. The limit is then $\lim_{x\rightarrow - 5}\frac{x - 3}{x - 5}$.

Step3: Substitute $x=-5$

Substitute $x=-5$ into $\frac{x - 3}{x - 5}$, we get $\frac{-5 - 3}{-5 - 5}=\frac{-8}{-10}=\frac{4}{5}$.

Answer:

$\frac{4}{5}$