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### Section 2.6 : Infinite Limits

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

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

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

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

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