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Link to solution. It is problem number 3.

$\lim_{(x,y) \to (0,0)}\frac{y^2 \sin^2 x}{x^4 + y^4}$

Solution: When $x$ is small [close to $0$], $\sin x$ has essentially the same growth as $x$. So the numerator is like $x^2y^2$, has the same degree as denominator. In this scenario, it's likely that the limit does not exist. To test this, we will try $y = mx$, where $m$ is just any real number. Then, we get $\frac{y^2 \sin^2 x}{x^4 + y^4} = \frac{m^2x^2 \sin^2 x}{(1 + m^4)x^4} = \frac{m^2 \sin^2 x}{(1 + m^4)x^2}$ Take the limit as $x$ goes to $0$, we will get $\frac{m^2}{1 + m^4}$ , which in particular means if we take $m = 2, m = 3$, the answers will not agree, and hence the limit does not exist.

I do not understand the last paragraph.

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    Oh, right. Thanks.2012-12-09

2 Answers 2

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$\frac{m^2\sin^2x}{(1+m^4)x^2}=\frac{m^2}{1+m^4}\left(\frac{\sin x}{x}\right)^2\xrightarrow [x\to 0]{}\frac{m^2}{1+m^4}\cdot 1^2=\frac{m^2}{1+m^4}$

Remember that

$\lim_{x\to 0}\frac{\sin x}{x}=1$

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For a limit of a single-variable function, the limit exists if the right- and left-hand limits exist and are equal.

In higher dimensions, we don't just have a right-hand and left-hand limit. We also have limits coming in from parabolas, from various lines, etc. For the limit in a multi-variable function to exist, the limit must exist and be equal from all possible paths.

In this problem, the author picked two different paths: $y=2x$ and $y=3x$. Evaluate these limits:

Path $y=2x$:

$\lim_{(x,y)\to(0,0)}\frac{y^2\sin^2x}{x^4+y^4}$ Substitute in $2x$ for $y$: $\lim_{x\to0}\frac{(2x)^2\sin^2x}{x^4+(2x)^4}$ Now this is just a Calc I limit. Skipping intermediate steps, we find the limit is equal to $\frac{4}{17}$.

Path $y=3x$:

$\lim_{(x,y)\to(0,0)}\frac{y^2\sin^2x}{x^4+y^4}$ Substitute in $3x$ for $y$: $\lim_{x\to0}\frac{(3x)^2\sin^2x}{x^4+(3x)^4}$ Now this is just a Calc I limit. Skipping intermediate steps, we find the limit is equal to $\frac{9}{82}$.

Therefore...

We know that for the limit to exist, the limits along all paths must be identical. We have a counterexample-the limit along the path $y=2x$ is not equal to the limit along the path $y=3x$. Thus, the limit doesn't exist.