$\lim_{(x,y)\to(0,0)} (y-\sin(y))/(x^2+y^2)$ $$
f(x,y) = \begin{cases} 0 &amp; \text{if } (x,y)=(0,0), \\[6pt]
\dfrac{y-\sin y}{x^2+y^2} &amp; \text{otherwise.} \end{cases}
$$

is $\lim\limits_{(x,y)\to(0,0)} \dfrac{y-\sin y}{x^2+y^2} = 0 \text{ ?}$ 

according to wolfarm no but can someone show me why?
I'm not familiar with functional analysis, but this question seems to be just a typical limit.

We will show the limit is $0$.

Observe that
$$\left|\frac{y-\sin y}{x^2+y^2}\right|=\frac{\left|y-\sin y\right|}{x^2+y^2}\le\frac{\left|y-\sin y\right|}{y^2}$$ so it is enough to prove that $\lim_{y\to 0}\frac{\left|y-\sin y\right|}{y^2}=0$. This can be done with the well known inequality $\frac{t^3}{3!}\ge t-\sin t$ for $t\ge 0$, or $\frac{|t|^3}{3!}\ge |t-\sin t|$ for all $t\in\mathbb{R}$.
Now, One of the clasical method for obtaining the limit of function with two variables in $(0,0)$ is by this fact that you can 
assume $x=r \, \cos(\theta)$ and $y=r \, \sin(\theta)$, then calculate the limit with new values for $x$ and $y$ when $r \to 0$, if after calculation limit, the final values of limit depend on to $\theta$, we conclude that, the function dose not have limit in $(0,0)$ but if the final values of limit be independent of $\theta$, we say the function has limit in $(0,0)$ with value that we obtained. so we have

$$\lim_{(x,y) \to (0,0)}\,\dfrac { y-\sin(y) } {x^{2}+y^{2} }
=\lim_{(r) \to (0)}\frac{r\,\sin(\theta)-\sin(r\,\sin(\theta))}{r^2}
$$
$$
\lim_{(r) \to (0)}\frac{r\,\sin(\theta)-\sin(r\,\sin(\theta))}{r^2}
=\lim_{(r) \to (0)}\frac{\frac{1}{3!}r^3\sin^3(\theta)}{r^2}=
\lim_{(r) \to (0)}\frac{1}{6}r\sin^3(\theta)=0
$$
that means your limit exist in $(0,0)$.