Prove $\lim_{(x,y)\rightarrow (0,3)}\frac{\tan(xy)}{x}=3$ exists Prove this limit exists:
$$\lim_{(x,y)\rightarrow (0,3)}\frac{\tan(xy)}{x}=3$$
My work:
Let $\epsilon >0$, $\delta=\cdots$
If $\sqrt{(x-0)^2+(y-3)^3}<\delta$
Then
\begin{align}
& \left|\frac{\tan(xy)}{x}-3\right|=\left|\frac{\tan(xy)-3x}{x}\right| = \left|\frac{\sin(xy)-3x\cos(xy)}{x\cos(xy)}\right| \\[10pt]
= {} & \frac{|\sin(xy)-3x\cos(xy)|}{|x\cos(xy)|} \leq \frac{|\sin(xy)|+3|x||\cos(xy)|}{|x||\cos(xy)|}\leq \frac{1+3|x|}{|x|}
\end{align}
In this step I'm stuck. Can someone help me?

A not so good solution (which can be made better of course) is 
\begin{align*}
\lim_{(x,y)\rightarrow(0,3)}\dfrac{\tan(xy)}{x}&=\lim_{(x,y)\rightarrow(0,3)}\dfrac{1}{x}\left(xy+\dfrac{1}{3}(xy)^{3}+\cdots\right)\\
&=\lim_{(x,y)\rightarrow(0,3)}\left(y+\dfrac{1}{3}x^{2}y^{3}+\cdots\right)\\
&=3.
\end{align*}
If we decide to go with $\epsilon$-argument, a painstaking way would be:
Observe that $\tan u\leq u+\dfrac{1}{3}u^{3}+\dfrac{1}{15}u^{5}$ for all real numbers $u$, just tackle with the function $\varphi(u)=u+3^{-1}u^{3}+15^{-1}u^{5}-\tan u$ and we know that $\varphi'(u)>0$ for $u\ne 0$ by the inequality follows by Mean Value Theorem.
On the other hand, by the similar fashion, we could have $\tan u\geq u+\dfrac{1}{3}u^{3}$ for $u\geq 0$. 
Given $\epsilon>0$ for all $|x|^{2}+|y-3|^{2}<\epsilon^{2}$ and $2<y<4$, one see that for $x>0$
\begin{align*}
\frac{\tan(xy)}{x}-3&\leq\frac{xy+3^{-1}(xy)^{3}+5^{-1}(xy)^{5}-3x}{x}\\
&=(y-3)+\frac{1}{3}x^{2}y^{3}+\frac{1}{5}x^{4}y^{5}\\
&\leq\epsilon+\frac{64}{3}\epsilon^{2}+\frac{512}{3}\epsilon^{4}.
\end{align*}
For $x<0$, one do the trick with $\dfrac{\tan(xy)}{x}=\dfrac{\tan(-xy)}{-x}$ and still get the upper bound.
Now we move on to the lower bound. For $x>0$, we have
\begin{align*}
\frac{\tan(xy)}{x}-3\geq\frac{xy+3^{-1}(xy)^{3}-3x}{x}=(y-3)+\frac{1}{3}x^{2}y^{3}\geq-\epsilon+\frac{8}{3}x^{2}\geq-\epsilon,
\end{align*}
the rest should be easy.
