Calculus I - Infinite Limits

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Section 2.6 : Infinite Limits

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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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$\mathop {\lim }\limits_{x \to {3^{\, - }}} f\left( x \right)$ 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 $x = 3$ .

$\mathop {\lim }\limits_{x \to {3^{\, + }}} f\left( x \right)$ 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.

$\mathop {\lim }\limits_{x \to 3} f\left( x \right)$ 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.