Calculus I - Infinite Limits

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Section 2.6 : Infinite Limits

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2. For $\displaystyle h\left( t \right) = \frac{{2t}}{{6 + t}}$ evaluate,

$\mathop {\lim }\limits_{t \to \, - {6^{\, - }}} h\left( t \right)$
$\mathop {\lim }\limits_{t \to \, - {6^{\, + }}} h\left( t \right)$
$\mathop {\lim }\limits_{t \to \, - 6} h\left( t \right)$

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$\mathop {\lim }\limits_{t \to \, - {6^{\, - }}} h\left( t \right)$ Show Solution

Let’s start off by acknowledging that for $t \to - {6^ - }$ we know $t < - 6$ .

For the numerator we can see that, in the limit, we will get -12.

The denominator takes a little more work. Clearly, in the limit, we have,

$6 + t \to 0$

but we can actually go a little farther. Because we know that $t < - 6$ we also know that,

$6 + t < 0$

More compactly, we can say that in the limit we will have,

$6 + t \to {0^ - }$

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

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{t \to \, - {6^{\, - }}} \frac{{2t}}{{6 + t}} = \infty }}$

Note that this also means that there is a vertical asymptote at $t = - 6$ .

$\mathop {\lim }\limits_{t \to \, - {6^{\, + }}} h\left( t \right)$ Show Solution

Let’s start off by acknowledging that for $t \to - {6^ + }$ we know $t > - 6$ .

For the numerator we can see that, in the limit, we will get -12.

The denominator will also work similarly to the first part. In the limit, we have,

$6 + t \to 0$

but we can actually go a little farther. Because we know that $t > - 6$ we also know that,

$6 + t > 0$

More compactly, we can say that in the limit we will have,

$6 + t \to {0^ + }$

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

$\require{bbox} \bbox[2pt,border:1px solid black]{{\mathop {\lim }\limits_{t \to \, - {6^{\, + }}} \frac{{2t}}{{6 + t}} = - \infty }}$

Note that this also means that there is a vertical asymptote at $t = - 6$ , which we already knew from the first part.

$\mathop {\lim }\limits_{t \to \, - 6} h\left( t \right)$ Show Solution

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

$\mathop {\lim }\limits_{t \to \, - {6^{\, - }}} h\left( t \right) \ne \mathop {\lim }\limits_{t \to \, - {6^{\, + }}} h\left( t \right)$

and so, from our basic limit properties we can see that $\mathop {\lim }\limits_{t \to \, - 6} h\left( t \right)$ does not exist.

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