Section 2.6 : Infinite Limits
4. For \(\displaystyle g\left( x \right) = \frac{{x + 7}}{{{x^2} - 4}}\) evaluate,
- \(\mathop {\lim }\limits_{x \to {2^{\, - }}} g\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to {2^{\, + }}} g\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to 2} g\left( x \right)\)
Show All Solutions Hide All Solutions
a \(\mathop {\lim }\limits_{x \to {2^{\, - }}} g\left( x \right)\) Show SolutionLet’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.