# How do you find lim sqrt(t^2+2)/(4t+2) as t->-oo?

Dec 25, 2016

#### Explanation:

For $t \ne 0$, we have $\sqrt{{t}^{2} + 2} = \sqrt{{t}^{2}} \sqrt{1 + \frac{2}{t} ^ 2}$ and $4 t + 2 = t \left(4 + \frac{2}{t}\right)$.

Furthermore, $\sqrt{{t}^{2}} = \left\mid t \right\mid$. In evaluating limit as $t \rightarrow - \infty$, we need only consider negative values for $t$. For $t < 0$, we get $\sqrt{{t}^{2}} = - t$

And ${\lim}_{t \rightarrow - \infty} \frac{c}{t} ^ n = 0$ for any number $c$ and any positive number $n$.

So,

${\lim}_{t \rightarrow - \infty} \frac{\sqrt{{t}^{2} + 2}}{4 t + 2} = {\lim}_{t \rightarrow - \infty} \frac{\sqrt{{t}^{2}} \sqrt{1 + \frac{2}{t} ^ 2}}{t \left(4 + \frac{2}{t}\right)}$

$= {\lim}_{t \rightarrow - \infty} \frac{- t \sqrt{1 + \frac{2}{t} ^ 2}}{t \left(4 + \frac{2}{t}\right)}$

$= {\lim}_{t \rightarrow - \infty} \frac{- \sqrt{1 + \frac{2}{t} ^ 2}}{4 + \frac{2}{t}}$

$= \frac{- \sqrt{1 + 0}}{4 + 0}$

$= \frac{- 1}{4}$