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### Section 2.6 : Infinite Limits

4. For $$\displaystyle g\left( x \right) = \frac{{x + 7}}{{{x^2} - 4}}$$ evaluate,

1. $$\mathop {\lim }\limits_{x \to {2^{\, - }}} g\left( x \right)$$
2. $$\mathop {\lim }\limits_{x \to {2^{\, + }}} g\left( x \right)$$
3. $$\mathop {\lim }\limits_{x \to 2} g\left( x \right)$$

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a $$\mathop {\lim }\limits_{x \to {2^{\, - }}} g\left( x \right)$$ Show Solution

Let’s start off by acknowledging that for $$x \to {2^ - }$$ we know $$x < 2$$.

For the numerator we can see that, in the limit, we will get 9.

Now let’s take care of the denominator. First, we know that if we square a number less than 2 (and greater than -2, which it is safe to assume we have here because we’re doing the limit) we will get a number that is less than 4 and so, in the limit, we will have,

${x^2} - 4 \to {0^ - }$

So, in the limit, the numerator is approaching a positive number and the denominator is an increasingly small negative number. The quotient must then be an increasing large negative number or,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to {2^{\, - }}} \frac{{x + 7}}{{{x^2} - 4}} = - \infty }}$

Note that this also means that there is a vertical asymptote at $$x = 2$$.

b $$\mathop {\lim }\limits_{x \to {2^{\, + }}} g\left( x \right)$$ Show Solution

Let’s start off by acknowledging that for $$x \to {2^ + }$$ we know $$x > 2$$.

For the numerator we can see that, in the limit, we will get 9.

Now let’s take care of the denominator. First, we know that if we square a number greater than 2 we will get a number that is greater than 4 and so, in the limit, we will have,

${x^2} - 4 \to {0^ + }$

So, in the limit, the numerator is approaching a positive number and the denominator is an increasingly small positive number. The quotient must then be an increasing large positive number or,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to {2^{\, + }}} \frac{{x + 7}}{{{x^2} - 4}} = \infty }}$

Note that this also means that there is a vertical asymptote at $$x = 2$$, which we already knew from the first part.

c $$\mathop {\lim }\limits_{x \to 2} g\left( x \right)$$ Show Solution

In this case we can see from the first two parts that,

$\mathop {\lim }\limits_{x \to {2^{\, - }}} g\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^{\, + }}} g\left( x \right)$

and so, from our basic limit properties we can see that $\mathop {\lim }\limits_{x \to 2} g\left( x \right)$ does not exist.

For the sake of completeness and to verify the answers for this problem here is a quick sketch of the function.

As we’re sure that you had already noticed there would be another vertical asymptote at $$x = - 2$$ for this function. For the practice you might want to make sure that you can also do the limits for that point.