Show $ f(x,y)= 2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - x\cos(\frac{1}{\sqrt{x^2 + y^2}}) \sqrt{x^2+y^2}^{-1}$ not continuous at $(0,0)$ $f(x,y) = 2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - \frac{x\cos(\frac{1}{\sqrt{x^2 + y^2}})}{ \sqrt{x^2+y^2}}
$
I'm a bit puzzled. The statement is obviously true if you plot the function.
Formal: $\quad f(a_n,b_n) \not\rightarrow (0,0)\quad for \quad (a_n,b_n) \rightarrow (0,0)$
I've tried something along the lines of $\sin(\frac{1}{a_n}) = \sin(\pi[4n+1]) = 1$, but this attempt remained fruitless.
 A: Try along two lines:
1) $x=0,y=t$, where $t \to 0$ Along this line the limit is zero.
2) $x=t,y=0$, $t \to 0$, along this line the limit doesn't exist, because limit $\lim_{t \to 0}\cos(\frac{1}{|t|})$ doesn't exist.
A: Let $x =r \cos(\phi), y = r\sin(\phi)$ then you have
\begin{align*}
\lim_{(x,y)\to(0,0)}f(x,y) &= \lim_{(x,y)\to(0,0)}2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - \frac{x\cos(\frac{1}{\sqrt{x^2 + y^2}})}{ \sqrt{x^2+y^2}} \\ 
&= \lim_{r\to 0}2 r\cos(\phi) \sin\left(\dfrac{1}{r}\right)- \dfrac{r\cos(\phi)\cos\left(\dfrac{1}{r}\right)}{r}
\end{align*}
Now $\lim_{r\to0}2r\cos(\phi)\sin\left(\frac{1}{r}\right)$ converges to $0$ indipendently from the angle $\phi$ but $\lim_{r\to 0} \cos(\phi)\cos\left(\frac{1}{r}\right)$ goes to $0$ dependently on the angle $\phi$ hence your function is not continuous in $(0,0)$
A: Technically, since this function isn't defined at $(0,0)$, the question should be worded “show that $f$ has no limit at $(0,0)$” or “show that there is no way to extend $f$ at $(0,0)$ so that the extension is continuous”.
Intuitively speaking, there are two reasons why a function might not have a limit: either the function wanders off altogether (at least along one approach path), or it oscillates between several points.¹ So to prove the absence of a limit, you can look for either of these properties:


*

*There is a sequence $(x_n,y_n)$ such that $\lim (x_n,y_n) = 0$ and $\lim |f(x_n,y_n)| = \infty$. It's sufficient to find a real parameter $t$ such that $\lim_{t\to 0} |f(t)| = \infty$. In other words, the function is locally unbounded at $(0,0)$.

*There are a sequences $(x_n,y_n)$ and $(x'_n,y'_n)$ such that $\lim (x_n,y_n) = \lim (x'_n,y'_n) = 0$ and $\lim f(x_n,y_n) \ne \lim f(x'_n,y'_n)$ (with both limits existing). Again, it's sufficient to find two real parametrizations leading to different limits, i.e. $\lim_{t\to 0} f(x(t),y(t)) \ne \lim_{t\to 0} f(x'(t),y'(t))$. In other words, there are two approach paths leading to different limit points.


These conditions are clearly sufficient to have no limit. They are in fact necessary — $\mathbb{R}^n$ is a locally compact metric space, so any sequence either wanders away to infinity (i.e. is unbounded) or has a limit point, and in the latter case it has a limit iff it has a single limit point. You don't need to know (or understand) that bit — all I'm saying here is that the technique above will always work if you find the right approach path(s).
It's clear that the function is bounded around $(0,0)$, because $\left| \dfrac{x}{\sqrt{x^2+y^2}} \right| \le 1$ so $|f(x,y)| \le 2|x| + 1$. So we need to look for two approach paths leading to different limit points.
$x=0$ is an obvious thing to try: $f(0,y) = 0$. Ok, $0$ is a limit point; all we need now is to find another.
Having tried $x=0$, let's try $y=0$. For $x \gt 0$, $f(x,0) = 2x \sin(1/x) - \cos(1/x)$. The first term has the limit $0$ (because $|\sin(1/x)| \le 1$ so $\lim_{x\to 0^+} x \sin(1/x) = 0$), so we need to prove that the second term does not have the limit $0$. This is intuitively clear if you graph it: $\cos(1/x)$ oscillates between $-1$ and $1$. We know that $\cos(1/x) = 1$ when $1/x = 2n\pi$ for some integer $n$; thus let's take $x_n = 1/(2n\pi)$. We have $\lim_{n\to\infty} f(x_n,0) = 0 - \lim_{n\to\infty} \cos(2n\pi) = 0 - \lim_{n\to\infty} 1 = -1$.
Since $f$ has both $0$ and $1$ as limit points at $(0,0)$, it does not have a limit.
