1. ## Limits

How can you show that the limit of this as $\displaystyle t_1$, $\displaystyle t_2$ approaches 0 is not 0...?

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

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

I'd really appreciate the help. Thanks.

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

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

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

I'd really appreciate the help. Thanks.
I think this is a crude way of saying

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

Is that right?

3. Yes, sorry if that wasn't clear. Any ideas?

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

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

Is that right?
Yes, but let's simplify this by letting $\displaystyle x=y$ since they both approach the same limit.

We now have:

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

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

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

We now have:

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

$\displaystyle \lim_{x \to 0}\frac{x^3}{1+x^6} = \frac{0}{1} = 0$
Now repeat with $\displaystyle y= x^{1/4}$ and let $\displaystyle x$ 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