# Need help proving limit does not exist!

1. Sep 20, 2012

### mathie_geek

1. The problem statement, all variables and given/known data
I need to prove a limit does not exist for f(x,y) as (x,y) -> (0,0). I already tried approaching from y=mx and I got the limit 0.

2. Relevant equations
f(x,y) = ((|x|^a)(|y|^b))/((|x|^c)+(|y|^d)) where a,b,c,d are positive numbers and a/c+b/d ≤
1

3. The attempt at a solution
I got limit = 0 approached from y=mx.
I can't use Squeez Theorem either. How should I approach this question? Just a hint would be good. Thanks!

Last edited: Sep 20, 2012
2. Sep 20, 2012

### SammyS

Staff Emeritus
Hello mathie_geek. Welcome to PF!

Edited:
[STRIKE]I assume your function really is:[/STRIKE]

I see you corrected it to be:

$\displaystyle f(x,y)=\frac{|x|^a\,|y|^b}{|x|^c+|y|^d}\ .$

Last edited: Sep 20, 2012
3. Sep 20, 2012

### mathie_geek

Thanks!
Trying to learn how to input functions :P
I missed a + so denominator is |x|^c + |y|^d

4. Sep 20, 2012

### SammyS

Staff Emeritus
Also, what is the restriction on a/c & b/d?

Is it as you said, $\displaystyle \frac{a}{c}+\frac{b}{d}<1\,,$

[STRIKE]or is it[/STRIKE] $\displaystyle \frac{a}{c}<1\ \text{ and }\ \frac{b}{d}<1\ ?$

Never mind, I see that it must be as originally stated.

Last edited: Sep 20, 2012
5. Sep 20, 2012

### mathie_geek

It's (a/c)+(b/d) <= 1
can anyone help

6. Sep 20, 2012

### jbunniii

Hint: try setting x = t^p, where p is some power, and y = t^q, where q is some other power.

7. Sep 20, 2012

### mathie_geek

I got lim y->0 ((|t|^ap)(|y|^b))/((|t|^pc)+|y|^d)) = 0
and lim x->0 ((|x|^a)(|t|^bq)/((|x|^c)+(|t|^dq)) = 0

Did I do something wrong?

8. Sep 20, 2012

### HallsofIvy

Staff Emeritus
Are there conditions on a, b, c, and d? If a+b is not larger than c+d you will NOT get "0" as limit for y=mx.

9. Sep 20, 2012

### jbunniii

No, I meant that you should pick appropriate values for p and q (in terms of a, b, c, and d) which will give you a counterexample. Hint: try to choose values which will get rid of the exponents in the denominator.

This will transform the problem into a limit involving a single variable, t. Then let that variable t approach zero. By doing this, x and y will approach (0,0) along a certain curve, instead of a straight line. If you do it right, you can get a nonzero limit.

10. Sep 20, 2012

### mathie_geek

ok, I tried x = t^(d/c) and y = t^(c/d) as two different limits so I got:
lim x->0 ((|x|^a)(|t|^bc/d))/((|x|^c)+(|t|^c))
those both =0 though.
I'm sorry I'm not that good at this.
Do you mean to put both of the substitutions into one limit of t->0?
a/c <= 1-(b/d)

11. Sep 20, 2012

### jbunniii

No, try to choose powers which will make the exponent in both terms of the denominator equal to 1.

Yes, make both substitutions at once, and then take the limit as t->0.

12. Sep 20, 2012

### SammyS

Staff Emeritus
m_geek,

Try letting (x, y) → (0, 0) along the path $\displaystyle y=x^{c/d}\ .$

That makes the denominator be fairly well behaved. I haven't checked it beyond that.

13. Sep 20, 2012

### mathie_geek

OH, I let y= t^(1/d) and x=t^(1/c)
so that lim t->0 (|t|^a/c+b/d-1)/2
and since a/c+b/d-1 <= 0,
then the limit would be 1/2 if a/c+b/d-1=0 or does not exist if a/c+b/d-1< 0.
since the two limits do not agree, limit d.n.e.
Is this correct? :D

14. Sep 20, 2012

### mathie_geek

I tried that but I don't know why it didn't work out for me because I kept getting 0 as the limit..

15. Sep 20, 2012

### SammyS

Staff Emeritus
Let's see your work for this, because it did work out for me & the limit is not zero.

16. Sep 20, 2012

### mathie_geek

I let $y = x^(c/d)$
so that it simplifies to $(|x|^ (a+(bc/d)))/2(|x|^c)$
which as x->0 limit becomes 0/2 ? I'm not sure how to take into account the inequality of a,b,c,d though

17. Sep 20, 2012

### SammyS

Staff Emeritus
I have fixed your Latex above, and repeat it below, so you can see the code.
$(|x|^{a+(bc/d)}) /(2(|x|^c))$

Now as a more readable fraction:

$\displaystyle \frac{|x|^{a+(bc/d)}}{2(|x|^c)}$

Write that expression using |x| only once.

Remember, $\displaystyle \frac{x^R}{x^T}=x^{R-T}\ .$

18. Sep 21, 2012

### mathie_geek

I got this after: $(|x|^{a+(bc/d)-c}) /{2}$
expanding the exponents I have $\displaystyle \frac{acd+bc^2-c^2d}{cd}$
and given $\displaystyle \frac{ad+bc-cd}{cd}<=0$ which is 1/c of the exponent..
So then |x| has an exponent negative so limit d.n.e right? cause limit of 1/|x| as x->0 d.n.e

19. Sep 21, 2012

### SammyS

Staff Emeritus
Actually, that's $\displaystyle \frac{ad+bc-cd}{d}\le 0$

What if $\displaystyle \frac{ad+bc-cd}{d}=0\ ?$

Now, maybe you can adjust the path so that the exponent is always zero.

20. Sep 21, 2012

### mathie_geek

Oh, I see. Since $\displaystyle \frac{ad+bc-cd}{d}> \displaystyle \frac{ad+bc-cd}{cd}$ then does that mean it HAS to = 0? If it's 0 then lim is 1/2