Prove Limit 0: f1(x,y)=4x³/3(x4+y4)2/3

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In summary, the conversation discusses finding f1 for (x,y)≠(0,0), proving that lim(x,y)→(0,0)f1(x,y)=0, and whether f1 is continuous at (0,0). The suggestion is to use polar coordinates to simplify the problem and find the limit along all paths, and references to differential equations are also provided.
  • #1
makavelian
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Homework Statement


let f(x,y)=(x4+y4)1/3
a)find f1 for (x,y)[tex]\neq[/tex](0,0)
b)prove that lim(x,y)[tex]\rightarrow[/tex](0,0)f1(x,y)=0
c) is f1 contintoius at (0,0)?

The Attempt at a Solution


f1=4x³/3(x4+y4)2/3
along any line y=mx, it is 0, along y=x² I run into a problem, where i get 4x1/3/3(1+x4)2/3, don't know what to do.

Once i pass this, I have a hunch i need to use squeeze theorm, yet I don't know how.

EDIT: realized that if I sub in 0 for x in the parabola limit, it goes to 0. I just need help organizing a squeeze theorem equivilent I am thinking [tex]\frac{4x^{3}}{3}[/tex]?
 
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  • #2
Unfortunately, knowing that the limit is 0 for all linear paths and even all parabolic paths is not enough! In order to be certain that the limit exists, and is 0, you would have to show that you get 0 as limit along all paths. And, of course, you can't do that by looking at different kinds of paths. There are simply too many.

I recommend changing to polar coordinates. That way, the distance to the origin depends on the single variable, r. If the limit, as r goes to 0, does not depend on [itex]\theta[/itex] that is the limit of the function as (x,y) goes to (0,0).
 
  • #3
so you're saying to make r=x²+y²? with x=rcos[tex]\vartheta[/tex] and y=rsin[tex]\vartheta[/tex]? how would I sub that in? or even isolate for x³?
 
  • #4
You just do the sub
[tex]
x=r\cos\theta,\ y=r\sin\theta:
[/tex]

[tex]
\frac{4x^3}{3(x^4+y^4)^{2/3}}\\
=\frac{4r^3\cos^3\theta}{3(r^4(\cos^4\theta+\sin^4\theta))^{2/3}}\\
=\frac{r^3}{r^{8/3}}\frac{4\cos^3\theta}{3(\cos^4\theta+\sin^4\theta)^{2/3}}\\
=r^{1/3}\frac{4\cos^3\theta}{3(\cos^4\theta+\sin^4\theta)^{2/3}}\\
[/tex]

It almost depends on the angle your approaching in, but yet it doesn't.
 
  • #5
[tex]\int\int_{-\infty}^{\infty} dx,dy=\infty\: dy +C[/tex] is the same as

[tex]\frac{\infty}{\infty}=\infty.[/tex] which is trivially true.

[tex]\int\int (x^4+y^4)^\frac{1}{3}=0 \lim\rightarrow{\infty}[/tex]

Thus:

[tex] \int\frac{4x^3}{3(x^4+y^4)^{2/3}}\ =0+C \lim\rightarrow\infty [/tex]

[tex]\int\int f(x,y)={(x4+y4)}^{1/3} dx,dy=0 \lim(x,y)\;\rightarrow{\infty}[/tex]

[tex](x,y)\neq 0;f1=1[/tex]

(0,0) is undefined.

[tex]\lim(x,y)\rightarrow (0,0)f1(x,y)=0[/tex]

Is an assymptote

[tex]\frac{d}{dx} f(x,y)={(x4+y4)}^{1/3=[/tex][tex]\rightarrow[/tex]

[tex]\frac{dx}{dy}\;\rightarrow[/tex]

[tex]=0 \lim\rightarrow\infty[/tex]

41e91672f57602d8b653dd5638351dfc.png


a=0

u=0

[tex]x,y\neq{0}[/tex]

Where the solution is an arbitrary constant or C or c=a.

see

http://en.wikipedia.org/wiki/Partial_differential_equation

becb2a10a4b16cbac0e667d492af8ec9.png


And

http://eqworld.ipmnet.ru/en/solutions/npde/npde1401.pdf

and

c65dd028c1c9fb80a8288ca893e949da.png
 
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  • #6
Thank you!
 

Related to Prove Limit 0: f1(x,y)=4x³/3(x4+y4)2/3

What is a limit in calculus?

A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a specific value. It helps us understand how a function behaves near a particular point on its graph.

How do you prove a limit in calculus?

To prove a limit, we must show that the values of the function get closer and closer to a specific value (known as the limit) as the input approaches a particular point. We can use algebraic techniques, such as factoring and simplifying, to manipulate the function and show that the limit exists and is equal to the desired value.

What is the limit of the given function?

The limit of the given function f1(x,y)=4x³/3(x4+y4)2/3 is 0. This can be determined by evaluating the function at the specific point (0,0) and showing that the values approach 0 as x and y approach 0.

What is the significance of proving a limit?

Proving a limit is important because it allows us to understand the behavior of a function near a specific point. It helps us make predictions about the function's values and behavior without actually evaluating the function at that point. This is crucial in many mathematical and scientific applications.

Are there any specific techniques or strategies for proving a limit?

Yes, there are several techniques and strategies for proving a limit, such as factoring, simplifying, substitution, and using special limit theorems. It is essential to carefully manipulate the function and use algebraic properties to show that the limit exists and is equal to the desired value.

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