the answer gave me 0/1 by dividing each term by t2
but its wrongs
Since you have a square root in the denominator, you will actually have to divide top and bottom by the SQUARE ROOT of the highest power.
In other words, multiply top and bottom by $\displaystyle \frac{1}{\sqrt{t^2}}$.
So $\displaystyle \lim_{t \to \infty}\frac{2t - 9}{\sqrt{t^2 + 8t + 5}} = \lim_{t \to \infty}\frac{(2t - 9)\frac{1}{\sqrt{t^2}}}{\sqrt{t^2 + 8t + 5}\left(\frac{1}{\sqrt{t^2}}\right)}$
$\displaystyle = \lim_{t \to \infty}\frac{2 - \frac{9}{t}}{\sqrt{1 + \frac{8}{t} + \frac{5}{t^2}}}$
$\displaystyle = \frac{2}{\sqrt{1}}$
$\displaystyle = 2$.
Hi wannous,
use $\displaystyle \sqrt{a}\sqrt{b}=\sqrt{ab}$
$\displaystyle \sqrt{t^2+8t+5}=\sqrt{t^2(1+\frac{8}{t}+\frac{5}{t ^2})}=\sqrt{t^2}\sqrt{1+\frac{8}{t}+\frac{5}{t^2}}$
$\displaystyle =t\sqrt{1+\frac{8}{t}+\frac{5}{t^2}}$
As t approaches infinity, the fractions go to zero.