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### Section 2.7 : Limits at Infinity, Part I

2. For $$h\left( t \right) = \sqrt[3]{t} + 12t - 2{t^2}$$ evaluate each of the following limits.

1. $$\mathop {\lim }\limits_{t \to \, - \infty } h\left( t \right)$$
2. $$\mathop {\lim }\limits_{t \to \,\infty } h\left( t \right)$$

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

To do this all we need to do is factor out the largest power of $$x$$ from the whole polynomial and then use basic limit properties along with Fact 1 from this section to evaluate the limit.

Note as well that we’ll convert the root over to a fractional exponent in order to allow it to be easier to deal with. Also note that this limit is a perfectly acceptable limit because the root is a cube root and we can take cube roots of negative numbers! We would only have run into problems had the index on the root been an even number.

\begin{align*}\mathop {\lim }\limits_{t \to \, - \infty } \left( {{t^{\frac{1}{3}}} + 12t - 2{t^2}} \right) & = \mathop {\lim }\limits_{t \to \, - \infty } \left[ {{t^2}\left( {\frac{1}{{{t^{\frac{5}{3}}}}} + \frac{{12}}{t} - 2} \right)} \right]\\ & = \left( {\mathop {\lim }\limits_{t \to \, - \infty } {t^2}} \right)\left[ {\mathop {\lim }\limits_{t \to \, - \infty } \left( {\frac{1}{{{t^{\frac{5}{3}}}}} + \frac{{12}}{t} - 2} \right)} \right] = \left( \infty \right)\left( { - 2} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty }}\end{align*}

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

For this part all of the mathematical manipulations we did in the first part did not depend upon the limit itself and so don’t need to be redone here. We can pick up the problem right before we actually took the limits and then proceed.

$\mathop {\lim }\limits_{t \to \,\infty } \left( {{t^{\frac{1}{3}}} + 12t - 2{t^2}} \right) = \left( {\mathop {\lim }\limits_{t \to \,\infty } {t^2}} \right)\left[ {\mathop {\lim }\limits_{t \to \,\infty } \left( {\frac{1}{{{t^{\frac{5}{3}}}}} + \frac{{12}}{t} - 2} \right)} \right] = \left( \infty \right)\left( { - 2} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \infty }}$