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Problem 58

# Show that $$\lim _{(x, y) \rightarrow(0,0)}\left[\left(2 x^{3}-y^{3}\right) /\left(x^{2}+y^{2}\right)=0\right.$$

Expert verified
In summary, to show that the given limit is 0, we first rewrote the function as $$f(x, y) = \frac{(2x - y)(x^2 + xy + y^2)}{x^2+y^2}$$ and analyzed the limit as both x and y approach 0. We observed that both the numerator and the denominator tend to 0 when x and y approach 0. By using polar coordinates (x = rcosθ and y = rsinθ) and simplifying the limit, we see that as r → 0, the overall function will also tend to 0. Therefore, we have shown that the given limit is indeed $$0: \lim _{(x, y) \rightarrow(0,0)}\frac{2 x^{3}-y^{3}}{x^{2}+y^{2}}=0$$.
See the step by step solution

## Step 1: Understanding the function

We are given a function in two variables: $$f(x, y) = \frac{2x^3 - y^3}{x^2+y^2}$$ We need to show that the limit of this function is 0 when both x and y tend to 0.

## Step 2: Simplifying the function

One possible simplification for this function is to rewrite the numerator as difference of cubes, that is: $$f(x, y) = \frac{(2x - y)(x^2 + xy + y^2)}{x^2+y^2}$$ Now, we can observe that (x^2+y^2) is a common factor in both the numerator and the denominator. However, we cannot directly cancel these factors, since the limit is evaluated as both x and y go to 0.

## Step 3: Analyzing the limit

We want to evaluate the following limit: $$\lim _{(x, y) \rightarrow(0,0)} \frac{(2x - y)(x^2 + xy + y^2)}{x^2+y^2}$$ One possible approach to evaluating this limit is to use the squeeze theorem. But first, let's analyze each term separately: In the numerator: - If x and y both tend to 0, then (2x - y) will also tend to 0 - Likewise, (x^2 + xy + y^2) will also tend to 0 when x and y tend to 0 In the denominator: - If x and y both tend to 0, (x^2+y^2) will also tend to 0 Now, let's rewrite the limit using polar coordinates by substituting x = rcosθ and y = rsinθ. $$\lim _{r \rightarrow 0} \frac{(2(rcos\theta) - rsin\theta)((rcos\theta)^2 + (rcos\theta)(rsin\theta) + (rsin\theta)^2)}{(rcos\theta)^2+(rsin\theta)^2}$$ Simplifying and canceling out the common terms: $$\lim _{r \rightarrow 0} \frac{(2cos\theta - sin\theta)(r^2(cos^2\theta + cos\theta sin\theta + sin^2\theta))}{r^2(cos^2\theta + sin^2\theta)}$$ It is now clear that as r → 0, the overall function will tend to 0 as well. Therefore, we have shown that the given limit is indeed 0: $$\lim _{(x, y) \rightarrow(0,0)}\left[\left(2 x^{3}-y^{3}\right)/\left(x^{2}+y^{2}\right)=0\right.$$

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