# How do you find lim 4t^2+6t+2 as t->-oo?

Dec 23, 2016

${\lim}_{t \to \text{-} \infty} \left(4 {t}^{2} + 6 t + 2\right) = + \infty$.

#### Explanation:

We see that we can't simply equate this to the sum of the limits of all 3 terms, since the first term shoots off to $+ \infty$, while the second term approaches $- \infty$. Different terms are pulling us towards different infinities. So what can we do?

Find the highest order of $t$ in the polynomial, and factor it out from the whole thing:

$\textcolor{w h i t e}{=} {\lim}_{t \to \text{-} \infty} \left(4 {t}^{2} + 6 t + 2\right)$

$= {\lim}_{t \to \text{-} \infty} {t}^{2} \left(4 + \frac{6}{t} + \frac{2}{t} ^ 2\right)$

Then, use "the limit of a product is the product of the limits":

$= \left[{\lim}_{t \to \text{-"oo)t^2] times [lim_(t->"-} \infty} \left(4 + \frac{6}{t} + \frac{2}{t} ^ 2\right)\right]$

Note: this rule is acceptable as long as we don't have one limit approaching $\pm \infty$ while the other one goes to $0$.

The first limit will approach $+ \infty$. The second will approach $4$. This is because dividing by larger and larger (negative) numbers makes the terms $\frac{6}{t}$ and $\frac{2}{t} ^ 2$ get closer to $0$, meaning their contributions to the sum $4 + \frac{6}{t} + \frac{2}{t} ^ 2$ will diminish as $t$ grows.

So we end up with

lim_(t->"-"oo)(4t^2+6t+2)=[lim_(t->"-"oo)t^2] times [lim_(t->"-"oo)(4+6/t+2/t^2)]

$\textcolor{w h i t e}{{\lim}_{t \to \text{-} \infty} \left(4 {t}^{2} + 6 t + 2\right)} = \left[\infty\right] \times \left[4\right]$
$\textcolor{w h i t e}{{\lim}_{t \to \text{-} \infty} \left(4 {t}^{2} + 6 t + 2\right)} = \infty$

Different terms may be pulling us toward different infinities, but the term with the highest power of $t$ is going to have the strongest "pull" as $t$ gets closer to $- \infty$. For ${\lim}_{x \to \pm \infty}$ of simple polynomials, we only need to keep the term with the highest power of $x$. In other words:
${\lim}_{t \to \text{-"oo)(4t^2+6t+2)=lim_(t->"-} \infty} 4 {t}^{2}$
$\textcolor{w h i t e}{{\lim}_{t \to \text{-} \infty} \left(4 {t}^{2} + 6 t + 2\right)} = + \infty$.
(The polynomial $4 {t}^{2} + 6 t + 2$ is a parabola that opens up, so this answer makes sense.)