# How do you find lim t/sqrt(4t^2+1) as t->-oo?

Dec 12, 2016

$- \frac{1}{2}$ (See below)

#### Explanation:

For $t \ne 0$, we have

$\frac{t}{\sqrt{4 {t}^{2} + 1}} = \frac{t}{\sqrt{{t}^{2} \left(4 + \frac{1}{t} ^ 2\right)}}$

$= \frac{t}{\sqrt{{t}^{2}} \sqrt{4 + \frac{1}{t} ^ 2}}$

We know that $\sqrt{{t}^{2}} = \left\mid t \right\mid$, so for $t < 0$, we have

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

$= \frac{- 1}{\sqrt{4 + \frac{1}{t} ^ 2}}$

As $t \rightarrow \infty$, the ration $\frac{1}{t} ^ 2 \rightarrow 0$, so we get

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

$= \frac{- 1}{\sqrt{4 + 0}} = - \frac{1}{2}$