# Limits

• November 9th 2008, 05:37 PM
WWTL@WHL
Limits
How can you show that the limit of this as $t_1$, $t_2$ approaches 0 is not 0...?

$lim_{t \to 0} \left( \frac{t_1 t_2^4}{t_1^2 + t_2^8} \right)$

I understand why it doesn't approach 0 since $t_1^2 + t_2^8$ approaches 0 quicker than $t_1 t_2^4$, but how can it be shown more formally?

I'd really appreciate the help. Thanks.
• November 9th 2008, 06:37 PM
Mathstud28
Quote:

Originally Posted by WWTL@WHL
How can you show that the limit of this as $t_1$, $t_2$ approaches 0 is not 0...?

$lim_{t \to 0} \left( \frac{t_1 t_2^4}{t_1^2 + t_2^8} \right)$

I understand why it doesn't approach 0 since $t_1^2 + t_2^8$ approaches 0 quicker than $t_1 t_2^4$, but how can it be shown more formally?

I'd really appreciate the help. Thanks.

I think this is a crude way of saying

$\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^2+y^8}$

Is that right?
• November 9th 2008, 11:19 PM
WWTL@WHL
Yes, sorry if that wasn't clear. Any ideas? :)
• November 10th 2008, 04:48 AM
colby2152
Quote:

Originally Posted by Mathstud28
I think this is a crude way of saying

$\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^2+y^8}$

Is that right?

Yes, but let's simplify this by letting $x=y$ since they both approach the same limit.

We now have:

$\lim_{x \to 0}\frac{x^5}{x^2+x^8}$

$\lim_{x \to 0}\frac{x^3}{1+x^6} = \frac{0}{1} = 0$
• November 10th 2008, 05:26 AM
CaptainBlack
Quote:

Originally Posted by colby2152
Yes, but let's simplify this by letting $x=y$ since they both approach the same limit.

We now have:

$\lim_{x \to 0}\frac{x^5}{x^2+x^8}$

$\lim_{x \to 0}\frac{x^3}{1+x^6} = \frac{0}{1} = 0$

Now repeat with $y= x^{1/4}$ and let $x$ go to zero from the right, since for a limit to exist it must give the same result however you approach the point in question.

CB