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### Section 2.4 : Limit Properties

7. Use the limit properties given in this section to compute the following limit. At each step clearly indicate the property being used. If it is not possible to compute any of the limits clearly explain why not.

$\mathop {\lim }\limits_{x \to \, - 5} \frac{{x + 7}}{{{x^2} + 3x - 10}}$
Hint : All we need to do is use the limit properties on the limit until we can use Properties 7, 8 and/or 9 from this section to compute the limit.
Show Solution
$\mathop {\lim }\limits_{x \to \, - 5} \frac{{x + 7}}{{{x^2} + 3x - 10}} = \frac{{\mathop {\lim }\limits_{x \to \, - 5} \left( {x + 7} \right)}}{{\mathop {\lim }\limits_{x \to \, - 5} \left( {{x^2} + 3x - 10} \right)}} \hspace{0.25in}{\mbox{Property 4}}$

Okay, at this point let’s step back a minute. We used property 4 here and we know that we can only do that if the limit of the denominator is not zero. So, let’s check that out and see what we get.

\begin{alignat*}{3}\mathop {\lim }\limits_{x \to \, - 5} \left( {{x^2} + 3x - 10} \right) & = \mathop {\lim }\limits_{x \to \, - 5} {x^2} + \mathop {\lim }\limits_{x \to \, - 5} 3x - \mathop {\lim }\limits_{x \to \, - 5} 10 & & \hspace{0.25in}{\mbox{Property 2}}\\ & = \mathop {\lim }\limits_{x \to \, - 5} {x^2} + 3\mathop {\lim }\limits_{x \to \, - 5} x - \mathop {\lim }\limits_{x \to \, - 5} 10 & & \hspace{0.25in}{\mbox{Property 1}}\\ & = {\left( { - 5} \right)^2} + 3\left( { - 5} \right) - 10 & & \hspace{0.25in}{\mbox{Properties 7, 8, & 9}}\\ & = 0 & & & \end{alignat*}

So, the limit of the denominator is zero so we couldn’t use property 4 in this case. Therefore, we cannot do this limit at this point (note that it will be possible to do this limit after the next section).