limit of (x*(y-1)^2*cosx)/(x^2+2(y-1)^2) as (x,y)->(0,1).

By evaluating along different paths this limit often goes to 0. This does not necessarily imply that it exists. So how would i prove that it exists. Can someone please show me how i can prove using delta - epsilon proof.

Thanks in advance.

To prove that the limit of a function exists using the delta-epsilon proof, you need to show that for any given epsilon (a small positive number), there exists a corresponding delta (a small positive number) such that whenever the distance between the input (x, y) and the limit point (0, 1) is less than delta, the distance between the function value and the limit value (0 in this case) is less than epsilon.

Here's a step-by-step guide on how to prove the limit using the delta-epsilon proof:

1. Begin by assuming ε > 0. We will prove that the limit of the given function as (x, y) approaches (0, 1) is 0, meaning the function value gets arbitrarily close to 0 as (x, y) gets arbitrarily close to (0, 1).

2. Use the expression of the function, f(x, y) = (x * (y - 1)^2 * cos(x)) / (x^2 + 2(y - 1)^2), and consider the distance between f(x, y) and 0.

3. We want to show that for any ε > 0, there exists a δ > 0 such that whenever 0 < sqrt(x^2 + (y - 1)^2) < δ, then |f(x, y)| < ε.

4. Start by factoring out the terms involving x and (y - 1) from the numerator and denominator of the expression:

f(x, y) = (x * (y - 1)^2 * cos(x)) / (x^2 + 2(y - 1)^2)
= (x * (y - 1)^2 * cos(x)) / x^2 * (1 + 2[(y - 1)/x]^2)

5. Observe that when (x, y) approaches (0, 1), (y - 1) approaches 0 and (y - 1)/x approaches 0/0 (an indeterminate form).

6. To handle the indeterminate form, manipulate the expression further by factoring out an extra x^2 in the denominator:

f(x, y) = (x * (y - 1)^2 * cos(x)) / x^2 * (1 + 2[(y - 1)/x]^2)
= (x * (y - 1)^2 * cos(x)) / x^2 * (1 + 2[(y - 1)/x]^2) * (x^2 / x^2)

Simplifying, we get:

f(x, y) = ((y - 1)^2 * cos(x)) / (1 + 2[(y - 1)/x]^2)

7. Now, consider the magnitude of f(x, y):

|f(x, y)| = |((y - 1)^2 * cos(x)) / (1 + 2[(y - 1)/x]^2)|

8. Note that |f(x, y)| ≤ (y - 1)^2 / (1 + 2[(y - 1)/x]^2) since |cos(x)| ≤ 1 for all x.

9. Since (y - 1)^2 ≥ 0 and (1 + 2[(y - 1)/x]^2) > 0, we can further simplify:

|f(x, y)| ≤ (y - 1)^2 / (1 + 2[(y - 1)/x]^2)
≤ (y - 1)^2 / 1

10. The expression becomes:

|f(x, y)| ≤ (y - 1)^2

11. Now we can see that as (x, y) approaches (0, 1), the term (y - 1)^2 approaches 0, and therefore, |f(x, y)| approaches 0 as well.

12. Let ε > 0 be given. We can choose δ = √ε. Then, for any 0 < √(x^2 + (y - 1)^2) < δ,

|f(x, y)| ≤ (y - 1)^2 < √ε^2 = ε

13. This shows that for any ε > 0, there exists δ > 0 such that whenever 0 < √(x^2 + (y - 1)^2) < δ, then |f(x, y)| < ε.

14. Therefore, by the delta-epsilon definition of a limit, the limit of the function as (x, y) approaches (0, 1) is 0.

By following these steps, you have proven using the delta-epsilon proof that the limit of the given function as (x, y) approaches (0, 1) exists and is equal to 0.

To prove the existence of the limit using a delta-epsilon proof, we need to show that for any given positive epsilon, there exists a positive delta such that the distance between the function and the limit is less than epsilon whenever the distance between the input point and the limit point is less than delta.

Let's proceed with the proof:

1. Start by assuming that the limit exists, and let epsilon be any positive number.

2. We need to find a delta such that if the distance between (x, y) and (0, 1) (i.e. ||(x, y) - (0, 1)||) is less than delta, then the distance between the function and the limit (0 in this case) is less than epsilon (i.e. |(x*(y-1)^2*cosx)/(x^2+2(y-1)^2) - 0| < epsilon).

3. Notice that the denominator in the function is always positive, so we can ignore it in the proof and focus on the numerator.

4. Rewrite the function as f(x, y) = x*(y-1)^2*cos(x). We want to show that |f(x, y) - 0| < epsilon.

5. Let's consider the absolute value of the function |f(x, y)| = |x*(y-1)^2*cos(x)|. Since cos(x) is bounded, we can find a constant M such that |cos(x)| <= M for all x.

6. Rewrite the absolute value |f(x, y)| <= |x*(y-1)^2*M|. Now, let's focus on the factor |(y-1)^2*M|. Since |y-1| can be small, we can choose a constant B such that |(y-1)^2*M| <= B for all y near 1.

7. Let's rewrite the absolute value |f(x, y)| <= |x|*B. Now, we only need to consider the factor |x|. To make |x| small, we need to choose a delta that restricts the possible values of x.

8. Choose delta = epsilon/B. Now, if ||(x, y) - (0, 1)|| < delta, then |x| <= delta = epsilon/B.

9. Thus, |f(x, y)| <= |x|*B <= (epsilon/B)*B = epsilon.

10. This shows that for any epsilon, there exists a delta such that if ||(x, y) - (0, 1)|| < delta, then |f(x, y)| < epsilon.

11. Therefore, by delta-epsilon definition, the limit of the function as (x, y) approaches (0, 1) is indeed 0.

This completes the proof of the existence of the limit using a delta-epsilon argument.