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Chapter 12: Problem 60

Use the method of your choice to evaluate the following limits. $$\lim _{(x, y) \rightarrow(0,0)} \frac{y^{2}}{x^{8}+y^{2}}$$

Short Answer

Expert verified
Answer: The limit of the function as both \(x\) and \(y\) approach \(0\) is \(0\).

Step by step solution

01

Rewrite the function as a limit equation

The function is given by: $$\lim _{(x, y) \rightarrow(0,0)} \frac{y^{2}}{x^{8}+y^{2}}$$ We will keep this form throughout the solution.
02

Check if the function is indeterminate

Before proceeding, let's check if the function is indeterminate when both \(x\) and \(y\) approach \(0\). Plugging in \(x=0\) and \(y=0\) into the function, we get: $$\frac{0^2}{0^8+0^2} = \frac{0}{0}$$ This is an indeterminate form, so we need to apply some method to find the limit.
03

Apply Polar coordinates

Since we have to evaluate the limit as both \(x\) and \(y\) approach \(0\), we can make use of polar coordinates to simplify the function. We can convert \(x\) and \(y\) to polar coordinates using the following formulas: $$x = r\cos(\theta)$$ $$y = r\sin(\theta)$$ Substituting these equations in the original function, we get: $$\lim _{(r, \theta) \rightarrow(0,0)} \frac{(r\sin(\theta))^{2}}{(r\cos(\theta))^8+(r\sin(\theta))^2}$$ Simplifying, we get: $$\lim _{(r, \theta) \rightarrow(0,0)} \frac{r^2\sin^2(\theta)}{r^8\cos^8(\theta) + r^2\sin^2(\theta)}$$
04

Simplify the function using polar coordinates

Now, we can separate out the terms containing \(r\) to simplify the function further: $$\lim _{(r, \theta) \rightarrow(0,0)} \frac{r^2\sin^2(\theta)}{r^2(\cos^8(\theta)+\sin^2(\theta))}$$ Cancel out \(r^2\) from both numerator and denominator: $$\lim _{(r, \theta) \rightarrow(0,0)} \frac{\sin^2(\theta)}{\cos^8(\theta) + \sin^2(\theta)}$$
05

Evaluate the limit

Now, we can evaluate the final limit as \(r\) approaches \(0\). Note that \(\theta\) can vary between \(0\) and \(2\pi\). No matter the value of \(\theta\), as \(r\) approaches \(0\), the value of the function becomes: $$\lim_{(r, \theta)\rightarrow(0, 0)} \frac{\sin^2(\theta)}{\cos^8(\theta) + \sin^2(\theta)} = 0$$ This is because the \(\sin^2(\theta)\) and \(\cos^8(\theta)\) terms are always finite, non-negative and \(\sin^2(\theta)\) term is in the numerator. Hence, the limit exists and is equal to \(0\).

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