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### Section 2.5 : Computing Limits

8. Evaluate $$\displaystyle \mathop {\lim }\limits_{x \to \, - 3} \frac{{\sqrt {2x + 22} - 4}}{{x + 3}}$$, if it exists.

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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. Simply factoring will not do us much good here so in this case it looks like we’ll need to rationalize the numerator.

\begin{align*}\mathop {\lim }\limits_{x \to \, - 3} \frac{{\sqrt {2x + 22} - 4}}{{x + 3}} & = \mathop {\lim }\limits_{x \to \, - 3} \frac{{\left( {\sqrt {2x + 22} - 4} \right)}}{{\left( {x + 3} \right)}}\frac{{\left( {\sqrt {2x + 22} + 4} \right)}}{{\left( {\sqrt {2x + 22} + 4} \right)}} = \mathop {\lim }\limits_{x \to \, - 3} \frac{{2x + 22 - 16}}{{\left( {x + 3} \right)\left( {\sqrt {2x + 22} + 4} \right)}}\\ & = \mathop {\lim }\limits_{x \to \, - 3} \frac{{2\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {\sqrt {2x + 22} + 4} \right)}} = \mathop {\lim }\limits_{x \to \, - 3} \frac{2}{{\sqrt {2x + 22} + 4}} = \frac{2}{8} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{4}}}\end{align*}