There is not really a lot to this problem. Simply recall the basic ideas for computing limits that we looked at in this section. In this case we see that if we plug in the value we get 0/0. Recall that this DOES NOT mean that the limit doesn’t exist. We’ll need to do some more work before we make that conclusion. All we need to do here is some simplification and then we’ll reach a point where we can plug in the value.

\[\mathop {\lim }\limits_{x \to \, - 5} \frac{{{x^2} - 25}}{{{x^2} + 2x - 15}} = \mathop {\lim }\limits_{x \to \, - 5} \frac{{\left( {x - 5} \right)\left( {x + 5} \right)}}{{\left( {x - 3} \right)\left( {x + 5} \right)}} = \mathop {\lim }\limits_{x \to \, - 5} \frac{{x - 5}}{{x - 3}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{5}{4}}}\]