How do you prove that the limit of # (x^3*y^2)/(x^2+y^2)# as (x,y) approaches (0,0) using the epsilon delta proof?

Answer 1

See explanation

If we have a function #f(x)# of one variable, then:
#lim_(x->c) f(x) = a" "# if and only if:
#AA epsilon > 0 EE delta > 0 : AA x in (c - delta, c + delta)" " abs(f(x) - a) < epsilon#
This can be generalised to a function from #RR^2 -> RR# as follows:
#lim_((x, y) -> (c, d)) f(x, y) = a" "# if and only if:
#AA epsilon > 0 EE delta > 0 : AA x in (c - delta, c + delta) AA y in (d - delta, d + delta)" " abs(f(x, y) - a) < epsilon#

In our example:

#f(x, y) = (x^3y^2)/(x^2+y^2)#
#(c, d) = (0, 0)#

and we will prove that:

#lim_((x, y) -> (0, 0)) = 0#
Given any #epsilon > 0#

Let:

#delta = { (sqrt(2) " if " epsilon > sqrt(2)), (epsilon " if " epsilon <= sqrt(2)) :}#

Let:

#x in (0 - delta, 0 + delta)#
#y in (0 - delta, 0 + delta)#

Then:

#x^2 < delta^2 <= 2#
#y^2 < delta^2 <= 2#

So:

#1/x^2+1/y^2 > 1/2+1/2 = 1#

So:

#1/(1/x^2+1/y^2) < 1#

Then:

If #x = 0# and #y = 0# then #f(x, y)# is undefined. So #(0, 0)# is not part of the domain of #f(x, y)#
If #x = 0# and #y != 0# or #x != 0# and #y = 0# then #f(x, y) = 0# and:
#abs(f(x, y) - 0) = 0 < epsilon#

Otherwise:

#abs(f(x, y) - 0) = abs((x^3y^2)/(x^2+y^2))#
#color(white)(abs(f(x, y) - 0)) = abs(x)(x^2y^2)/(x^2+y^2)#
#color(white)(abs(f(x, y) - 0)) = abs(x)(1)/(1/y^2+1/x^2)#
#color(white)(abs(f(x, y) - 0)) < abs(x) < delta < epsilon#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To prove the limit of the function (x^3*y^2)/(x^2+y^2) as (x,y) approaches (0,0) using the epsilon-delta proof, we need to show that for any given epsilon greater than zero, there exists a delta greater than zero such that whenever the distance between (x,y) and (0,0) is less than delta, the distance between the function and its limit (which is 0) is less than epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a delta > 0 such that whenever 0 < sqrt(x^2 + y^2) < delta, we have |(x^3*y^2)/(x^2+y^2) - 0| < epsilon.

Now, let's simplify the expression |(x^3*y^2)/(x^2+y^2) - 0|:

|(x^3y^2)/(x^2+y^2) - 0| = |x^3y^2/(x^2+y^2)| = |x^3*y^2|/|x^2+y^2|

Since x^2 + y^2 > 0, we can rewrite the expression as:

|x^3*y^2|/(x^2+y^2)

Now, we can use the triangle inequality to further simplify:

|x^3y^2|/(x^2+y^2) ≤ (|x^3||y^2|)/(x^2+y^2) ≤ (|x|^3*|y|^2)/(x^2+y^2)

Since |x| and |y| are both less than or equal to sqrt(x^2 + y^2), we can substitute:

(|x|^3*|y|^2)/(x^2+y^2) ≤ (sqrt(x^2 + y^2)^3*sqrt(x^2 + y^2)^2)/(x^2+y^2)

Simplifying further:

(sqrt(x^2 + y^2)^3*sqrt(x^2 + y^2)^2)/(x^2+y^2) = (x^2 + y^2)^(5/2)/(x^2+y^2) = (x^2 + y^2)^(3/2)

Now, we can see that (x^2 + y^2)^(3/2) is equal to the distance between (x,y) and (0,0), which is sqrt(x^2 + y^2). Therefore, we have:

|(x^3*y^2)/(x^2+y^2) - 0| ≤ (x^2 + y^2)^(3/2) = sqrt(x^2 + y^2)

Now, we can choose delta = epsilon. If 0 < sqrt(x^2 + y^2) < delta = epsilon, then we have:

|(x^3*y^2)/(x^2+y^2) - 0| ≤ sqrt(x^2 + y^2) < delta = epsilon

Thus, we have shown that for any given epsilon > 0, there exists a delta > 0 such that whenever 0 < sqrt(x^2 + y^2) < delta, we have |(x^3y^2)/(x^2+y^2) - 0| < epsilon. Therefore, the limit of the function (x^3y^2)/(x^2+y^2) as (x,y) approaches (0,0) is indeed 0.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7