Showing Limit of xy^3/(x^2+y^2) = 0 using Definition

[TheAntithesis]

Homework Statement

Apply the definition of the limit to show that

lim (x,y)-->(0,0) xy^3/(x^2+y^2) = 0

Homework Equations

Definition of the limit:
lim (x,y)-->(a,b) f(x,y) = L if for every number epsilon > 0 there is a corresponding number delta > 0 such that if (x,y) is in the domain and 0 < sqrt((x-a)^2 + (y-b)^2) < delta then |f(x,y) - L| < epsilon

The Attempt at a Solution

So far I've just plugged in the numbers:

Let epsilon > 0. We want to find delta > 0 such that
if 0 < sqrt(x^2 + y^2) < delta then |xy^3/(x^2 + y^2) - 0| < epsilon

I have no idea what to do next.
Note that it says to show that this is the limit using the definition given above.

 

[Quinzio]

Is not true that the limit is 0.
If you let [itex]y=0[/itex] and [itex]x \to 0[/itex], we have the
[tex]\lim_{x \to 0} \frac{1}{x} \ne 0[/tex]

 

[TheAntithesis]

If y = 0 then xy^3 = 0 then xy^3/(x^2+y^2) = 0/x^2 = 0 so the limit must be 0

 

[NeroKid]

u can't take limit of the numerator without taking limit of the denominator so u basically wrong

 

[TheAntithesis]

Wolfram?
http://www.wolframalpha.com/input/?i=lim+%28x%2Cy%29-%3E%280%2C0%29+xy^3%2F%28x^2+%2B+y^2%29

 

[NeroKid]

sry my bad :D

 

[Quinzio]

If y = 0 then xy^3 = 0 then xy^3/(x^2+y^2) = 0/x^2 = 0 so the limit must be 0

Oh yes, sure.

 

[HallsofIvy]

I would switch to polar coordinates- that way the "distance to (0, 0)" is given by the single variable, r.

In polar coordinates, the function is
[tex]\frac{(r cos(\theta)(r sin(\theta))^3}{r^2}= r^2 cos(\theta)sin^3(\theta)[/tex]

Of course, [itex]|cos(\theta)sin^3(\theta)|\le 1[/itex] for all [itex]\theta[/itex].

 

[TheAntithesis]

Ok let me know if this makes any sense at all...

After changing it to polar coordinates, we get [itex] \lim_{r \to 0} r^2 cos(\theta)sin^3(\theta)[/itex]

Let [itex]\epsilon > 0[/itex]. We want to find [itex]\delta > 0[/itex] such that

if [itex] 0<r< \delta[/itex] then [itex]|r^2 cos(\theta)sin^3(\theta) - 0| < \epsilon [/itex]

since [itex]|cos(\theta)sin^3(\theta)| \le 1[/itex] for all [itex]\theta[/itex]

[itex]r^2 cos(\theta)sin^3(\theta) \le r^2[/itex]

Thus if we choose [itex]\delta^2 = \epsilon[/itex] and let [itex]0<r<\delta[/itex] then

[itex]|r^2 cos(\theta)sin^3(\theta) - 0| \le r^2 \le \delta^2 = \epsilon[/itex]

And hence [itex] \lim_{(x,y) \to (0,0)} = 0 [/itex]