Section 2.6 : Infinite Limits
1. For \(\displaystyle f\left( x \right) = \frac{9}{{{{\left( {x - 3} \right)}^5}}}\) evaluate,
- \(\mathop {\lim }\limits_{x \to {3^{\, - }}} f\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to {3^{\, + }}} f\left( x \right)\)
- \(\mathop {\lim }\limits_{x \to 3} f\left( x \right)\)
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a \(\mathop {\lim }\limits_{x \to {3^{\, - }}} f\left( x \right)\) Show Solution
Let’s start off by acknowledging that for \(x \to {3^ - }\) we know \(x < 3\).
For the numerator we can see that, in the limit, it will just be 9.
The denominator takes a little more work. Clearly, in the limit, we have,
\[x - 3 \to 0\]but we can actually go a little farther. Because we know that \(x < 3\) we also know that,
\[x - 3 < 0\]More compactly, we can say that in the limit we will have,
\[x - 3 \to {0^ - }\]Raising this to the fifth power will not change this behavior and so, in the limit, the denominator will be,
\[{\left( {x - 3} \right)^5} \to {0^ - }\]We can now do the limit of the function. In the limit, the numerator is a fixed positive constant and the denominator is an increasingly small negative number. In the limit, the quotient must then be an increasing large negative number or,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to {3^{\, - }}} \frac{9}{{{{\left( {x - 3} \right)}^5}}} = - \infty }}\]Note that this also means that there is a vertical asymptote at \(x = 3\).
b \(\mathop {\lim }\limits_{x \to {3^{\, + }}} f\left( x \right)\) Show Solution
Let’s start off by acknowledging that for \(x \to {3^ + }\) we know \(x > 3\).
As in the first part the numerator, in the limit, it will just be 9.
The denominator will also work similarly to the first part. In the limit, we have,
\[x - 3 \to 0\]and because we know that \(x > 3\) we also know that,
\[x - 3 > 0\]More compactly, we can say that in the limit we will have,
\[x - 3 \to {0^ + }\]Raising this to the fifth power will not change this behavior and so, in the limit, the denominator will be,
\[{\left( {x - 3} \right)^5} \to {0^ + }\]We can now do the limit of the function. In the limit, the numerator is a fixed positive constant and the denominator is an increasingly small positive number. In the limit, the quotient must then be an increasing large positive number or,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{x \to {3^{\, + }}} \frac{9}{{{{\left( {x - 3} \right)}^5}}} = \infty }}\]Note that this also means that there is a vertical asymptote at \(x = 3\), which we already knew from the first part.
c \(\mathop {\lim }\limits_{x \to 3} f\left( x \right)\) Show Solution
In this case we can see from the first two parts that,
\[\mathop {\lim }\limits_{x \to {3^{\, - }}} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {3^{\, + }}} f\left( x \right)\]and so, from our basic limit properties we can see that \(\mathop {\lim }\limits_{x \to 3} f\left( x \right)\) does not exist.
For the sake of completeness and to verify the answers for this problem here is a quick sketch of the function.
