Why does $\lim_{x\to 0} \frac {\sin (xy)}{x} \to y $? Let $f(x,y) = \frac{\sin (xy)}{x}$ for $x\neq 0$. How should you define $f(0,y)$ for $y\in \mathbb{R}$ so as to make $f$ a continuous function on all of $\mathbb{R}^2$?
So in order for a function to be continuous the limit of the function approaching the point has to equal the value of the function at the point. In this case the trouble point is $(0,\,y)$. Now i looked at some other posts and took the suggestion of using L'Hopital's rule. Using it I get the right solution, but I don't understand why doing L'Hopital does provide the right solution? If i am in a multivariate environment, how am i allowed to use something like L'Hopital? 
 A: From calculus of one variable it is known that $$\lim_{t\to 0}\frac{\sin t}{t}=1$$
Then, if $y\neq 0$ let $t=xy$, $x=t/y$, we have
\begin{align*}
\lim_{x\to 0}\frac{\sin (xy)}{x}&=\lim_{x\to 0}\left[\frac{y}{1}\frac{\sin (xy)}{xy}\right]\\
&=y\lim_{t\to 0}\frac{\sin t}{t}\\
&=y(1)\\
&=y
\end{align*}
Then, in order to get $f$ be continuous in $\mathbb{R}^2$, it is needed $f(0,y)=y$
A: In a multivariate environment, you can't guarantee that the answer you get using L'Hopital will make your function continuous. However, you can guarantee that if your function can be made continuous, L'Hopital's rule will give you the right answer.
In your case, you want $f(x,y)$ to be continuous on $\mathbf{R}^2$. That means, in particular, if you hold $y$ constant and vary $x$, then $f(x,y)$ must be continuous as a function of $x$ for all possible $y$. To make this condition hold, you use L'Hopital's rule to fill in the discontinuity, viewing $f(x,y)$ as a function of just one variable, $x$.
This does NOT guarantee your function is continuous everywhere. Once you get your values for $f(0,y)$, you have to go back and PROVE these values make your function continuous. However, if it's possible for your function to be continuous, these will necessarily be the correct values.
A: You can use the fundamental limit for $\sin$ instead: $$\frac{\sin(xy)}{x} = y\frac{\sin(xy)}{xy} \stackrel{x \to 0}{\longrightarrow} y \cdot 1 = y.$$The point is that $y$ is fixed here.
