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From my multivariable textbook:

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

Wolfram indicates that the limit DNE, but does not list the steps used to solve.

Is there a particular substitution that I'm overlooking?

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up vote 7 down vote accepted

Calculate the limits along the lines $y=x$ and $y=2x$, as $(x,y)\to (0,0)$, these are unequal.

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As a procedural matter, how would one recognize the need to compare the limit when y=x against the limit when y=2x? – nitrl Mar 1 '14 at 22:06
$\frac{y^2\sin^2 x}{x^4+y^4} = \frac{x^2y^2}{x^4+y^4}\frac{\sin^2x}{x^2}$, now as $x\to 0$, we have $\frac{\sin^2x}{x^2}\to 1$. Again one writes $\frac{x^2y^2}{x^4+y^4}=\frac{1}{\frac{x^2}{y^2}+\frac{y^2}{x^2}}$, that gives the idea of trying to evaluate the limits along two lines with different slopes $y=m_1x$ and $y=m_2x$, so that the limit values become different. – r9m Mar 1 '14 at 22:13

You can look at the function on the line $y=x$. Then you look for $$lim_{x\to0}\frac{x^2\sin^2 x}{2x^4}$$. Since $sin x$ is equivalent in $0$ to $x$, you then see that this limit is $1/2$. Then look at the line $y=2x$, and the limit is now $4/17$ if i'm not mistaken. So the limit of the function in (0,0) does not exist.

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Thanks. I believe the second limit evaluates to 0, however. See:… – nitrl Mar 2 '14 at 22:08
You're welcome. I think it should be $$lim_{x\to0}\frac{4x^2\sin^2 x}{17x^4}$$ in your formula, so the limit is 4/17 indeed… – Asinus Mar 4 '14 at 20:19
Ah, understood. – nitrl Mar 5 '14 at 1:14

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