# Two Variable Limits

• Jan 27th 2010, 11:06 AM
matt.qmar
Two Variable Limits
Hello!
Trying to find these limits (or show they do not exist)
Using the two path method (to show non-existance)
Not sure how to show they exist (unless they go a certain number by an identity?)
Any help appriciated! Thanks!

$\displaystyle \lim_{(x,y)-->(0,0)} \frac{\sin{x^3y^3}}{x^6+y^6}$

$\displaystyle \lim_{(x,y)-->(2,-3)} \frac{x\sqrt{y}}{\sqrt{x^3-y^3}}$

$\displaystyle \lim_{(x,y)-->(1,2)} \frac{y-x}{\sqrt{x}(y-2)}$
• Jan 27th 2010, 04:20 PM
Aryth
You can say that these limits do not exist in similar ways to single variable limits.

The first one yields an indeterminant form... So it may or may not exist.

To tackle this one is a bit of work but I'll work on it a little bit and edit this later.

For the second one, look below.

The third one is a classic example of multiple paths to prove nonexistence.

If you plug everything in you immediately see a 0 in the denominator, so it is indeterminate. So, we try a random path... Say, y = x... Then we get:

$\displaystyle \lim_{(x,y) \to (1,2)} \frac{y - x}{\sqrt{x}(y-2)} \$$\displaystyle = \lim_{(x,y) \to (1,2)} \frac{0}{\sqrt{x}(x-2)} = 0$

Now, let's try $\displaystyle x = y^2$ (Just to mix it up)

$\displaystyle = \lim_{(x,y) \to (1,2)} \frac{y(1 - y)}{y(y^2 - 2)} \$ = $\displaystyle -\lim_{(x,y) \to (1,2)} \frac{y - 1}{y^2 - 2}$

We plug and chug to get:

$\displaystyle \lim_{(x,y) \to (1,2)} \frac{y-x}{\sqrt{x}(y - 2)} = -\frac{1}{2}$

Which is NOT zero, so it does not exist.

Hope that helps.
• Jan 27th 2010, 04:37 PM
pickslides
Quote:

Originally Posted by Aryth
The second one actually exists, but it is misleading because you can sense that an i is going to come out... And that can't be a limit... But upon inspection you can actually solve it using the plug and chug method, plugging everything in you'll get:

$\displaystyle \lim_{(x,y) \to (2,-3)} \frac{x\sqrt{y}}{\sqrt{x^3 - y^3}} = \frac{2\sqrt{-3}}{\sqrt{-5}}$

If you look closely the i's cancel, leaving:

Just to clarifiy

$\displaystyle \lim_{(x,y) \to (2,-3)} \frac{x\sqrt{y}}{\sqrt{x^3 - y^3}} = \frac{2\sqrt{-3}}{\sqrt{(2)^3 - (-3)^3}} = \frac{2\sqrt{-3}}{\sqrt{8 - -27}} = \frac{2\sqrt{-3}}{\sqrt{35}} \neq \frac{2\sqrt{-3}}{\sqrt{-5}}$
• Jan 27th 2010, 04:39 PM
Aryth
Quote:

Originally Posted by pickslides
Just to clarifiy

$\displaystyle \lim_{(x,y) \to (2,-3)} \frac{x\sqrt{y}}{\sqrt{x^3 - y^3}} = \frac{2\sqrt{-3}}{\sqrt{(2)^3 - (-3)^3}} = \frac{2\sqrt{-3}}{\sqrt{8 - -27}} = \frac{2\sqrt{-3}}{\sqrt{35}} \neq \frac{2\sqrt{-3}}{\sqrt{-5}}$

You're absolutely right... I forgot it was cubed and squared them out of instinct... I apologize.
• Jan 27th 2010, 05:43 PM
matt.qmar
Thanks!!

For the first limit, so far, I have, if we approach from any line y=mx,

$\displaystyle \lim_{(x,y)-->(0,0)} \frac{\sin{(x^3(mx)^3)}}{x^6+(mx)^6}$

$\displaystyle \lim_{(x,y)-->(0,0)} \frac{\sin{(x^6m^3)}}{(m+1)^6x^6}$

$\displaystyle \frac{1}{(m+1)^6} * \lim_{(x,y)-->(0,0)} \frac{\sin{x^6m^3}}{x^6}$

as $\displaystyle x^6$ goes to 0, $\displaystyle m^3x^6$ goes to 0. So the limit of

$\displaystyle \lim_{(x,y)-->(0,0)} \frac{\sin{x^6m^3}}{x^6}$

is 1, and the whole thing is,

$\displaystyle \frac{1}{(m+1)^6}$

which is different for each m, so the limit does not exist.

Not sure if this is legitimate? Hope so! Is there another way to do it?
• Jan 27th 2010, 05:51 PM
Aryth
Quote:

Originally Posted by matt.qmar
Thanks!!

For the first limit, so far, I have, if we approach from any line y=mx,

$\displaystyle \lim_{(x,y)-->(0,0)} \frac{\sin{(x^3(mx)^3)}}{x^6+(mx)^6}$

$\displaystyle \lim_{(x,y)-->(0,0)} \frac{\sin{(x^6m^3)}}{(m+1)^6x^6}$

$\displaystyle \frac{1}{(m+1)^6} * \lim_{(x,y)-->(0,0)} \frac{\sin{x^6m^3}}{x^6}$

as $\displaystyle x^6$ goes to 0, $\displaystyle m^3x^6$ goes to 0. So the limit of

$\displaystyle \lim_{(x,y)-->(0,0)} \frac{\sin{x^6m^3}}{x^6}$

is 1, and the whole thing is,

$\displaystyle \frac{1}{(m+1)^6}$

which is different for each m, so the limit does not exist.

Not sure if this is legitimate? Hope so! Is there another way to do it?

Just one thing:

$\displaystyle \lim_{(x,y) \to (0,0)} \frac{\sin{(x^6m^3)}}{x^6} = m^3$

NOT 1.