Showing $\lim_{(x,y) \to (0,0)} \sin (xy) / xy = 1$

Solution 1:

If you know the Taylor series for sin(z) then you can expand it for $z=x \cdot y$. Then divide by $x \cdot y$

Solution 2:

Show first that $\lim_{x\to0}\lim_{y\to0} \dfrac{\sin xy}{xy}$ is $1$, then show that $\lim_{y\to0}\lim_{x\to0} \dfrac{\sin xy}{xy}$ is $1$, then show that if $x$ and $y$ are functions that depends of a parameter $t$ where $lim_{t\to t_0} x(t) = lim_{t\to t_0} y(t) = 0$ then $\lim_{t\to t_0} \dfrac{\sin (x(t)y(t))}{x(t)y(t)} = 1$ for any kind of parametrization.

first you have to know that $$\lim_{x\to 0}\dfrac{\sin x}{x} = \lim_{x\to 0}\dfrac{\cos x}{1} = 1$$ (I used L'Hopital)

Let's do the first:

$$\lim_{x\to0}\lim_{y\to0} \dfrac{\sin xy}{xy} = \lim_{x\to 0}\lim_{y\to0} \dfrac{x\cos{xy}}{x} = \lim_{x\to 0}\lim_{y\to0}\dfrac{\cos{xy}}{1} = \lim_{x\to 0}1 = 1$$

Let's do the second :

$$\lim_{y\to0}\lim_{x\to0} \dfrac{\sin xy}{xy} = \lim_{y\to 0}\lim_{x\to0} \dfrac{y\cos{xy}}{y} = \lim_{y\to 0}\lim_{x\to0}\dfrac{\cos{xy}}{1} = \lim_{y\to 0}1 = 1$$

For the third part we do a little trick, we declare that $x(t)$ and $y(t)$ are differentiables on the neighborhood of $t_0$ and $x(t_0)=y(t_0)=0$ so:

$$\lim_{t\to t_0} \dfrac{\sin (x(t)y(t))}{x(t)y(t)}= \lim_{t\to t_0} \dfrac{\cos (x(t)y(t))(x'(t)y(t)+y'(t)x(t))}{x'(t)y(t)+y'(t)x(t)}=\lim_{t\to t_0}\cos (x(t)y(t)) =\cos(0\cdot0)=1$$