Evaluate the limit

**wannous** · April 17th 2010, 08:29 AM

the answer gave me 0/1 by dividing each term by t2
but its wrongs

**Prove It** · April 17th 2010, 08:42 AM

Originally Posted by wannous

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 $\frac{1}{\sqrt{t^2}}$ .

So $\lim_{t \to \infty}\frac{2t - 9}{\sqrt{t^2 + 8t + 5}} = \lim_{t \to \infty}\frac{(2t - 9)\frac{1}{\sqrt{t^2}}}{\sqrt{t^2 + 8t + 5}\left(\frac{1}{\sqrt{t^2}}\right)}$

$= \lim_{t \to \infty}\frac{2 - \frac{9}{t}}{\sqrt{1 + \frac{8}{t} + \frac{5}{t^2}}}$

$= \frac{2}{\sqrt{1}}$

$= 2$ .

**wannous** · April 17th 2010, 08:47 AM

clear explanation, thank you my good sir

**Archie Meade** · April 17th 2010, 09:00 AM

Originally Posted by wannous

the answer gave me 0/1 by dividing each term by t2
but its wrongs

Hi wannous,

use $\sqrt{a}\sqrt{b}=\sqrt{ab}$

$\sqrt{t^2+8t+5}=\sqrt{t^2(1+\frac{8}{t}+\frac{5}{t ^2})}=\sqrt{t^2}\sqrt{1+\frac{8}{t}+\frac{5}{t^2}}$

$=t\sqrt{1+\frac{8}{t}+\frac{5}{t^2}}$

As t approaches infinity, the fractions go to zero.

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