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I have the function $f:\mathbb{R}^2\rightarrow\mathbb{R}\ f(x,y) = \left\{\begin{matrix} \sin\frac{x^3y}{x^4+y^4}, & (x,y) \in \mathbb{R}^2 \setminus\{(0,0)\}\\ 0, & (x,y) = (0,0). \end{matrix}\right.$
I need to study the continuity of the function $f$.

I tried to calculate the limits as $(x,y) \rightarrow (0,0)$ of $f$. So $\lim_{(x,y)\rightarrow(0,0)}\sin\frac{x^3y}{x^4+y^4} = \sin(0) = 0$ So this would imply that $f$ is continuous at $(0,0)$, right?

user26857
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Raducu Mihai
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3 Answers3

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HINT:

Note that along $x=0$, $\sin\left(\frac{x^3y}{x^4+y^4}\right)=0$. Along $x=y$, $\sin\left(\frac{x^3y}{x^4+y^4}\right)=\sin(1/2)$. What can you can conclude?

Mark Viola
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  • Honestly, nothing. I just started multivariable limits and I can't see a solution from your hint – Raducu Mihai Feb 09 '19 at 21:43
  • But you know limits with one variable, right? If we have a function of one variable then there is a limit if and only if both one sided limits exist and are equal to each other. In multivariable limits it is the same thing, just there are infinitely many one sided limits instead of two. But still, if two one sided limits are not equal then there can't be a limit. – Mark Feb 09 '19 at 21:45
  • Ok, so the limit does not exist, so the function is not continuous, right? So in order to solve these limits I need to evaluate both limits (first when $x\rightarrow 0 $ and second when $y\rightarrow 0$? – Raducu Mihai Feb 09 '19 at 21:48
  • @RaducuMihai The limits along $x=0$ and $y=0$ are identical. Choose any other path that is not along a coordinate axis. – Mark Viola Feb 09 '19 at 21:49
  • No, it's not that simple. As I said there are infinitely many one sided limits, not just two, so it is usually tough to prove a limit exists this way. But if you can show two one sided limits are different from each other then there is no limit. So for example, if you take $(x,y)\to(0,0)$ along the path $(0,t)$ then the limit will be $0$. If you take along the path $(t,t)$ then the limit is $\sin(\frac{1}{2})$. Hence there are two different one sided limits. The function is not continuous at the origin. – Mark Feb 09 '19 at 21:51
  • Ok I think I got it. I need to evaluate the limit for two different paths. In your example the path are $(0,y) and (x,x)$, right? Because the two limits are not equal implies that the limit does not exist, right? – Raducu Mihai Feb 09 '19 at 21:51
  • Yes, that's it. – Mark Feb 09 '19 at 21:51
  • Thank you so much! – Raducu Mihai Feb 09 '19 at 21:51
  • You're welcome! My pleasure. – Mark Viola Feb 09 '19 at 21:52
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I don't understand your solution. I'll give a hint on how to solve it. What limit do you get if you take $(x,y)\to (0,0)$ along the curve $(t,0)$? And how about the curve $(t,t)$?

Mark
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$h(x,y)=\frac{x^3y}{x^4+y^4}$ has no limit at $(0;0)$. Puting $x=r\cos{\theta}$ and $y=r\sin{\theta}$ yields: $f(x,y)=\frac{\sin{\theta}\cos^3{\theta}}{\cos^4{\theta}+\sin^4{\theta}}$, choosing $\theta=0$ and $\theta=\frac{\pi}{3}$ provide two different values.

user26857
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DINEDINE
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