# Explain this solution if it is correct.

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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It means the limit depends on what particular $\,m\,$ we take, which shows the limit doesn't exist as if it did then it'd be the same no matter how we approach $\,(0,0)\,$ – DonAntonio Dec 9 '12 at 2:39
But If we take the limit while x -> 0, won't that cause the whole thing go to 0/0 since sin^2 (0) = 0 in the numerator and (0)^2 = 0 in the denominator. What basic concept am I missing? – cRaZiRiCaN Dec 9 '12 at 2:43
look at the answers as this is way too long (for me) for a comment. – DonAntonio Dec 9 '12 at 2:44
You’re missing the fact that $$\frac{\sin^2x}{x^2}=\left(\frac{\sin x}x\right)^2\to 1^2=1$$ as $x\to 0$. – Brian M. Scott Dec 9 '12 at 2:44
Oh, right. Thanks. – cRaZiRiCaN Dec 9 '12 at 2:46

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

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.

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