# Thread: Limit of functions of 2 variables

1. ## Limit of functions of 2 variables

Find $\lim_{(x,y) \to (0,0)} \frac{3x^2y}{x^2+y^2}$ if it exists.

After testing different paths, the limit appears to be 0, so let's try an $\epsilon - \delta$ proof. I got a bit rusty at this, I just don't know how to finish it off lol.

We need to prove if for every $\epsilon > 0$ there exists a $\delta$ such that if $0 < \sqrt{x^2+y^2} < \delta$ then $\left|\frac{3x^2y}{x^2+y^2} - 0\right|< \epsilon$

We know that $x^2 \le x^2 + y^2 \implies 1 \le \frac{x^2+y^2}{x^2} \implies 1 \ge \frac{x^2}{x^2+y^2} \implies 3|y| \ge \frac{3|y|x^2}{x^2+y^2} \implies 3\sqrt{y^2} \ge \frac{3|y|x^2}{x^2+y^2}$

But $3\sqrt{x^2+y^2} \ge 3\sqrt{y^2} \implies \frac{3|y|x^2}{x^2+y^2} \le 3\sqrt{x^2+y^2}$

Thus we can take $\epsilon = 3\sqrt{x^2+y^2}$ but then how do we relate this back to $\delta$ again? I forgot lols

2. You could convert to polars...

Remember that $x = r\cos{\theta}, y = r\sin{\theta}, x^2 + y^2 = r^2$, then

$\lim_{(x, y) \to (0, 0)}\frac{3x^2y}{x^2 + y^2} = \lim_{r \to 0}\frac{3(r\cos{\theta})^2r\sin{\theta}}{r^2}$

$= \lim_{r \to 0}\frac{3r^3\cos^2{\theta}\sin{\theta}}{r^2}$

$= \lim_{r \to 0}3r\cos^2{\theta}\sin{\theta}$

$= 0$.

3. The best way to handle a problem like this is to change to polar coordinates. That way, only the single variable, r, measure how "close" to (0, 0) you are. If the limit as r goes to 0 exists and is the same for all $\theta$, then the limit itself exists.

Of course, $x= r cos(\theta)$, $y= r sin(\theta)$, and $x^2+ y^2= r^2$ so this problem becomes

$\lim_{r\to 0}\frac{3(r cos(\theta))^2(r sin(\theta))}{r^2}$
$\lim_{r\to 0}3 cos^2(\theta)sin(\theta)\frac{r^3}{r^2}= 3 cos^2(\theta)sin(\theta)\lim_{r\to 0} r= 0$
for all $\theta$.

If you want to do a strict " $\epsilon-\delta$" proof just do it as you would for one variable with r and leave $\theta$ alone.

(You can't take " $\epsilon= \sqrt{x^2+ y^2}$" for two reasons:
1) $\epsilon$ is given- it can be anything. It is $\delta$ you are to "take".
2) You must get $\delta$ in terms of $\epsilon$, not x and y. It must be a constant, not a function of x and y.)

4. Beat you :P

5. Usagi, you would take [LaTeX ERROR: Convert failed] and I think you'll be correct.

Halls, Prove, here is an example from my textbook that may trouble you

Let [LaTeX ERROR: Convert failed]

With your method, set $x = r \cos(t)$ and $y = r \sin(t)$. Then the new function is

$F(r,t) = \frac{r^3\cos(t)\sin(t)}{r^2(\cos^2(t)+r^2\sin^4(t ))} = \frac{r \cos(t)\sin(t)}{\cos^2(t)+r^2\sin^4(t)}$

Taking the limit as r approaches 0, we get 0.

HOWEVER, let $x = y^2[/tex] in [tex]f(x,y)$. Now we have

$f(y^2,y) = \frac{y^4}{2y^4} = 1/2$ for all values.

Hence the limit does not exist and your method fails!

6. Thank you all!! I understand now, thanks once again!!

7. The sandwich theorem works too:

$\left| \dfrac{3x^2y}{x^2+y^2} \right| \leq \dfrac{3x^2y}{x^2} = 3 y$

8. Originally Posted by Vlasev
Usagi, you would take [LaTeX ERROR: Convert failed] and I think you'll be correct.

Halls, Prove, here is an example from my textbook that may trouble you

Let [LaTeX ERROR: Convert failed]

With your method, set $x = r \cos(t)$ and $y = r \sin(t)$. Then the new function is

$F(r,t) = \frac{r^3\cos(t)\sin(t)}{r^2(\cos^2(t)+r^2\sin^4(t ))} = \frac{r \cos(t)\sin(t)}{\cos^2(t)+r^2\sin^4(t)}$

Taking the limit as r approaches 0, we get 0.

HOWEVER, let $x = y^2$ in [LaTeX ERROR: Convert failed] . Now we have

$f(y^2,y) = \frac{y^4}{2y^4} = 1/2$ for all values.

Hence the limit does not exist and your method fails!

Hi Vlasev,

we can't use this method here, but anyway, there is no reason to apply it here, since there is not any "x²+y²", or "xy"...No one has said that it worked all the time, i think

9. I think it is misleading to just supply a method and not mention its eventual limitation. I can see the OP trying to apply it on more tricky limits and get the wrong answer. Worse, I see the OP using it as a "proof" when in fact it is not.

10. Originally Posted by HallsofIvy
That way, only the single variable, r, measure how "close" to (0, 0) you are. If the limit as r goes to 0 exists and is the same for all $\theta$, then the limit itself exists.
Oh, I hadn't seen this before my last post...This sentence is clearly false.
The method that consists in changing to polar coordinates is good in a lot of cases, but theta must not be fixed.

11. Originally Posted by Vlasev
... here is an example from my textbook that may trouble you

Let [LaTeX ERROR: Convert failed]

With your method, set $x = r \cos(t)$ and $y = r \sin(t)$. Then the new function is

$F(r,t) = \frac{r^3\cos(t)\sin(t)}{r^2(\cos^2(t)+r^2\sin^4(t ))} = \frac{r \cos(t)\sin(t)}{\cos^2(t)+r^2\sin^4(t)}$

Taking the limit as r approaches 0, we get 0.

HOWEVER, let $x = y^2$ in [LaTeX ERROR: Convert failed] . Now we have

$f(y^2,y) = \frac{y^4}{2y^4} = 1/2$ for all values.

Hence the limit does not exist and your method fails!
The method works correctly if it is correctly used!... in this case the limit...

$\displaystyle \lim_{r \rightarrow 0} \frac{r\ \cos t \ \sin(t)}{\cos^{2} t+r^{2}\ \sin^4(t)}$

... exist if and only if it is the same for any t function of r. That is verified in 'almost all' the cases and the limit is 0... but not if is...

$\displaystyle \cos t = r\ \sin^{2} t$

... where the limit is $\frac{1}{2}$ ...

Kind regards

$\chi$ $\sigma$