# Limits

• Oct 15th 2008, 01:37 PM
WWTL@WHL
Limits
Having trouble on these two questions. I have no idea and help would be much appreciated. Thanks.

Find
$\displaystyle \lim_{t \to 3} \frac{t^2 - 2t - 3}{t^2 + 4t - 21}$

Find
$\displaystyle \lim_{t \to \infty} \sqrt{t^2 + t} - t$
• Oct 15th 2008, 01:44 PM
Moo
Hello,
Quote:

Originally Posted by WWTL@WHL
Having trouble on these two questions. I have no idea and help would be much appreciated. Thanks.

Find
$\displaystyle \lim_{t \to 3} \frac{t^2 - 2t - 3}{t^2 + 4t - 21}$

Factorise !
$\displaystyle t^2-2t-3=(t-3)(t+1)$

and $\displaystyle t^2+4t-21=(t-7)(t+3)$

Quote:

Find
$\displaystyle \lim_{t \to \infty} \sqrt{t^2 + t} - t$
Multiply and divide by its conjugate, that is $\displaystyle \frac{\sqrt{t^2+t}{\color{red}+}t}{\sqrt{t^2+t}+t}$
• Oct 15th 2008, 01:57 PM
WWTL@WHL
Thanks for the quick reply, Moo. :)

Quote:

Originally Posted by Moo
Hello,

Factorise !
$\displaystyle t^2-2t-3=(t-3)(t+1)$

and $\displaystyle t^2+4t-21=(t-7)(t+3)$

Sorry, I'm still have trouble with this. So substituting 3 in for t, and you have 0 on the numerator, and -3 on the denominator, which suggests it'll tend to 0, as t approaches 3?

But $\displaystyle t^2+4t-21 = 0$ when t=3 as well. So then you'd have 0/0 which is undefined?

Quote:

Multiply and divide by its conjugate, that is $\displaystyle \frac{\sqrt{t^2+t}{\color{red}+}t}{\sqrt{t^2+t}+t}$
Excellent. Thanks a lot. It tends to 1/2.
• Oct 15th 2008, 02:03 PM
Moo
Quote:

Originally Posted by WWTL@WHL
Sorry, I'm still have trouble with this. So substituting 3 in for t, and you have 0 on the numerator, and -3 on the denominator, which suggests it'll tend to 0, as t approaches 3?

But $\displaystyle t^2+4t-21 = 0$ when t=3 as well. So then you'd have 0/0 which is undefined?

Excuse me (Bow)

It's actually $\displaystyle t^2+4t-21=(t-3)(t+7)$

Do you see any simplification ? :)
• Oct 15th 2008, 02:09 PM
WWTL@WHL
Thanks. :)

$\displaystyle \lim_{t \to 3} \frac{t^2 - 2t - 3}{t^2 + 4t - 21}$ = $\displaystyle \lim_{t \to 3} \frac{t+1}{t+7} = \frac{2}{5}$

Does that seem ok?
• Oct 15th 2008, 02:19 PM
Moo
Quote:

Originally Posted by WWTL@WHL
Thanks. :)

$\displaystyle \lim_{t \to 3} \frac{t^2 - 2t - 3}{t^2 + 4t - 21}$ = $\displaystyle \lim_{t \to 3} \frac{t+1}{t+7} = \frac{2}{5}$

Does that seem ok?

Yes :)

You should add a step :
$\displaystyle \lim_{t \to 3} \frac{t^2 - 2t - 3}{t^2 + 4t - 21}=\lim_{t \to 3} \frac{(t-3)(t+1)}{(t-3)(t+7)}=\ldots$

;)