# Need help proving limit does not exist!

mathie_geek

## Homework Statement

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.

## Homework 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

## 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!

edit: Sorry, missed a + sign in the denominator!

Last edited:

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## Homework Statement

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.

## Homework 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

## The Attempt at a Solution

I got limit = 0 approached from y=mx.
I can't use Squeeze Theorem either. How should I approach this question? Just a hint would be good. Thanks!
Hello mathie_geek. Welcome to PF!

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

I see you corrected it to be:

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

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

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

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

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

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

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?

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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.

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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?
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.

mathie_geek
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.
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?
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.
a/c <= 1-(b/d)

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ok, I tried x = t^(d/c) and y = t^(c/d)
No, try to choose powers which will make the exponent in both terms of the denominator equal to 1.

Do you mean to put both of the substitutions into one limit of t->0?
Yes, make both substitutions at once, and then take the limit as t->0.

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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.

mathie_geek
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.

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

mathie_geek
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.

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

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I tried that but I don't know why it didn't work out for me because I kept getting 0 as the limit..
Let's see your work for this, because it did work out for me & the limit is not zero.

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

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

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

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

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

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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
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.

mathie_geek
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.

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

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...
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..
...
Let's go back to the exponent:
$\displaystyle \frac{acd+bc^2-c^2d}{cd}=\frac{c(ad+bc-cd)}{cd}=\frac{ad+bc-cd}{d}\ .$​
I have no idea as to why you multiplied that by 1/c .

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
Unless you meant to say that "a,b,c,d are positive integers [STRIKE]numbers[/STRIKE]" (emphasizing integers, rather than numbers), the following inequality
$\displaystyle \frac{ad+bc-cd}{d}>\frac{ad+bc-cd}{cd} \ \ \text{ is not true.}$​
The number, c, might very well be less than 1 if a, b, c, and d are not required to be integers.

All that I said was, if a/c + b/d ≤ 1, then $\displaystyle \frac{ad+bc-cd}{d}\le 0$.

If you choose the case of it being an equality, (choose the = sign), then the limit is defined, with said limit being 1/2.

However, that's far from being the general case for this expression.