$\lim\limits_{(x,y)\rightarrow(0,0)} \frac{7 \sin(2 x) x^2 y}{2 x^3 + 2 x y^2}$ Does the following limit exist?
$$\lim_{(x,y)\rightarrow(0,0)}  \frac{7 \sin(2 x) x^2 y}{2 x^3 + 2 x y^2}$$
I've shown that it exist but i'm unsure about how to find the value of the limit.
 A: Hint: Note that
$$
\frac{7\sin(2x)x^2y}{2x^3+2xy^2}=\frac72\sin(2x)\frac{y/x}{1+y^2/x^2}
$$
Since $(t-1)^2\ge0$, for any $t\in\mathbb{R}$, $\left|\frac{t}{1+t^2}\right|\le\frac12$. Therefore,
$$
\left|\frac{7\sin(2x)x^2y}{2x^3+2xy^2}\right|\le\frac74|\sin(2x)|
$$
Note that you cannot approach on any path that intersects the $y$-axis since the function is undefined there.
A: Note that
$$\frac{7 \sin(2 x) x^2 y}{2 x^3 + 2 x y^2}=7\cdot\frac{\sin(2x)}{2x}\cdot\frac{x^2y}{x^2+y^2}.$$
When $(x,y)\to (0,0)$, we have $x\to 0$ and $r=\sqrt{x^2+y^2}\to 0$, therefore, we have 
$$\left|\frac{\sin(2x)}{2x}\right|\to 1\mbox{ as }x\to 0$$
and 
$$\left|\frac{x^2y}{x^2+y^2}\right|=\left|\frac{r^3\cos^2\theta\sin\theta}{r^2}\right|\leq r\to 0.$$
A: If you are sure that the limitation you will compute is exist then it must be unique. So we can find  the limitation by a special way in $x-O-y$ plane such as $y=kx$.Then we have:
\begin{equation}
\lim_{(x,y)\rightarrow (0,0)}\frac{7\sin(2x)x^2y}{2x^3+2xy^2}=\lim_{(x,y)\rightarrow (0,0)}\frac{7\sin(2x)x^2kx}{2x^3+2x^3k^2}=\lim_{(x,y)\rightarrow (0,0)}\frac{7\sin(2x)}{2+2k^2}=0
\end{equation}
But I will point out this method is valid if you check the existence of the limitation.
Alternatively, you can use the equivalent infinitesimal technique. 
\begin{equation}
\lim_{(x,y)\rightarrow (0,0)}\frac{7\sin(2x)x^2y}{2x^3+2xy^2}=\lim_{(x,y)\rightarrow (0,0)}\frac{14x^3y}{2x^3+2xy^2}=0
\end{equation}
A: I'm going to be a little persnickety here.  The OP's function is of the form
$${f(x,y)\over xg(x,y)}$$
Therefore, technically it does not have a limit at $(0,0)$ because, as defined, it has no values along the $y$-axis.  This was mentioned in robjohn's answer, but it's implication was ignored:  In most texts, the existence of a limit is predicated on the function being defined at all points within some open ball (minus the point at which the limit is sought).
Of course this is arguably a silly objection because the OP's function is really of the form
$$xf(x,y)\over xg(x,y)$$
and the $x$'s can "clearly" be canceled, leaving something that is defined at all points other than $(0,0)$.  But what's "clear" to the eye may or may not be clear, say to a mindless computer, which will happily compute numerator and denominator separately and then balk at dividing $0$ into $0$.
