Question

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

Answer 1

Please see below.

Explanation:

For `t != 0`, we have `sqrt(t^2+2) = sqrt(t^2)sqrt(1+2/t^2)` and `4t+2 = t(4+2/t)`.

Furthermore, `sqrt(t^2) = abst`. In evaluating limit as `t rarr -oo`, we need only consider negative values for `t`. For `t < 0`, we get `sqrt(t^2) = -t`

And `lim_(trarr-oo)c/t^n = 0` for any number `c` and any positive number `n`.

So,

`lim_(trarr-oo)sqrt(t^2+2)/(4t+2) = lim_(trarr-oo)(sqrt(t^2) sqrt(1+2/t^2))/(t(4+2/t))`

`= lim_(trarr-oo)(-tsqrt(1+2/t^2))/(t(4+2/t))`

`= lim_(trarr-oo)(-sqrt(1+2/t^2))/(4+2/t)`

`= (-sqrt(1+0))/(4+0)`

`= (-1)/4`