Limit of $f(x,y)$ I have a hard time evaluating 
$$
\lim_{(x, y)\to(0, 0)} \frac{e^{xy}\sin(xy)}{xy}
$$
I keep getting the answer that the limit doesn't exist but actually the answer is $1$. Help.
 A: Note that $(x,y) \to 0$ if and only if $x^2 + y^2 \to 0$. This fact combined with the inequality $|xy| \leq (x^2 + y^2) / 2$ means that the following chain of implications holds:
$$(x,y) \to 0 \implies x^2 + y^2 \to 0 \implies xy \to 0$$
This justifies rewriting the problem by setting $u = xy$ and taking the limit as $u \to 0$:
$$\lim_{u \to 0}\frac{e^u \sin(u)}{u} = \left(\lim_{u \to 0}e^u\right) \left( \lim_{u \to 0}\frac{\sin(u)}{u}\right) = (1)(1) = 1$$
A: Recall from elementary geometry that the sine function satisfies the inequalities
$$x\cos(x)\le \sin(x)\le x \tag 1$$
for $|x|\le \pi/2$.  Then, we have from $(1)$
$$\cos(xy)\le \frac{\sin(xy)}{xy}\le 1 \tag2$$


In addition, I showed in This Answer using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities
$$1+x\le e^x\le \frac{1}{1-x}$$
for $x<1$


Putting $(2)$ and $(3)$ together, we have
$$(1+xy)\cos(xy) \le \frac{e^{xy}\sin(xy)}{xy}\le \frac{1}{1-xy}$$
whereupon application of the squeeze theorem reveals
$$\bbox[5px,border:2px solid #C0A000]{\lim_{(x,y)\to (0,0)}\frac{e^{xy}\sin(xy)}{xy}=1}$$
A: Simply use the Taylor series $$e^{xy} = 1 + xy + \frac{x^2y^2}{2} + \frac{x^3 y^3}{6} +\cdots$$ and $$\sin(xy) = xy - \frac{x^3y^3}{6} + \frac{x^5y^5}{120} \pm \cdots$$ and you'll end up with the representation $$\frac{e^{xy}\sin(xy)}{xy} = 1 + \mathcal O(xy)$$ which holds as $(x,y) \to (0,0)$.
A: You can try polar coordinates for any point where the function is defined.:
$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\implies \lim_{(x,y)\to(0,0)}e^{xy}\frac{\sin xy}{xy}=\lim_{r\to0}e^{r^2\cos t\sin t}\frac{\sin r^2cos t\sin t}{r^2\cos t\sin t}=e^0\cdot1=1$$
