1. Limits

Having trouble on these two questions. I have no idea and help would be much appreciated. Thanks.

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

Find
$\lim_{t \to \infty} \sqrt{t^2 + t} - t$

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

Find
$\lim_{t \to 3} \frac{t^2 - 2t - 3}{t^2 + 4t - 21}$
Factorise !
$t^2-2t-3=(t-3)(t+1)$

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

Find
$\lim_{t \to \infty} \sqrt{t^2 + t} - t$
Multiply and divide by its conjugate, that is $\frac{\sqrt{t^2+t}{\color{red}+}t}{\sqrt{t^2+t}+t}$

3. Thanks for the quick reply, Moo.

Originally Posted by Moo
Hello,

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

and $t^2+4t-21=(t-7)(t+3)$
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 $t^2+4t-21 = 0$ when t=3 as well. So then you'd have 0/0 which is undefined?

Multiply and divide by its conjugate, that is $\frac{\sqrt{t^2+t}{\color{red}+}t}{\sqrt{t^2+t}+t}$
Excellent. Thanks a lot. It tends to 1/2.

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

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

Do you see any simplification ?

5. Thanks.

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

Does that seem ok?

6. Originally Posted by WWTL@WHL
Thanks.

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

Does that seem ok?
Yes

You should add a step :
$
\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$