I'll appreciate it if someone can throw me in the right direction on this one:

prove:

Attachment 21189

Thanks

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- Mar 18th 2011, 08:35 AMSpinRatMulti-variable limit
I'll appreciate it if someone can throw me in the right direction on this one:

prove:

Attachment 21189

Thanks - Mar 18th 2011, 09:44 AMtopsquark
For the limit to exist it must be the same no matter which direction we approach it. I'd recommend approaching the origin along the lines y = x and y = -x.

y = x:

$\displaystyle \displaystyle \lim_{(x,y) \to (0,0)} \frac{sin(x^2)~sin^3(y)}{x^6 + y^4}$

$\displaystyle \displaystyle = \lim_{x \to 0} \frac{sin(x^2)~sin^3(x)}{x^6 + x^4}$

This is in the form to use l'Hopital's rule, but its a mess to do it this way. Expanding a bit:

$\displaystyle \displaystyle = \lim_{x \to 0} \left ( \frac{sin(x^2)}{x^2} \cdot \frac{sin(x)}{x} \cdot \frac{sin(x)}{x} \cdot \frac{sin(x)}{x^2 + 1} \right )$

$\displaystyle \displaystyle = \lim_{x \to 0} \frac{sin(x^2)}{x^2} \cdot \lim_{x \to 0} \frac{sin(x)}{x} \cdot \lim_{x \to 0} \frac{sin(x)}{x} \cdot \lim_{x \to 0} \frac{sin(x)}{x^2 + 1}$

$\displaystyle \displaystyle = 1 \cdot 1 \cdot 1 \cdot \lim_{x \to 0} \frac{sin(x)}{x^2 + 1}$

This last limit is 0. y = -x is done in the same way.

-Dan - Mar 18th 2011, 09:58 AMSpinRat
But that doesn't prove the limit.

Actually, in the first section of this question we were asked to show that for a different function the limit exist when approaching through__every__line {(x,y) | y=ax or x=0}, yet the limit doesn’t exist.

this was the other one:

Attachment 21190 - Mar 18th 2011, 10:07 AMgirdav
Use $\displaystyle |\sin t|\leq |t|$ and $\displaystyle ab\leq \frac {a^2+b^2}2$.

- Mar 19th 2011, 08:35 AMSpinRat
been at it all day, still no go. I think I'm missing something obvious. plz help...

Attachment 21198