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

Question

2617 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-12T17:25:42

$-1/2$ (See below)

For $t != 0$ , we have

$t/sqrt(4t^2+1) = t/sqrt(t^2(4+1/t^2))$

$= t/(sqrt(t^2)sqrt(4+1/t^2))$

We know that $sqrt(t^2) = abst$ , so for $t < 0$ , we have

$= t/(-t(sqrt4+1/t^2))$

$= (-1)/sqrt(4+1/t^2)$

As $trarroo$ , the ration $1/t^2 rarr0$ , so we get

$lim_(xrarr-oo) t/sqrt(4t^2+1) = lim_(xrarr-oo) (-1)/sqrt(4+1/t^2)$

$= (-1)/sqrt(4+0) = -1/2$

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