I am asked to find if there exists the limit of $\lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{y}$. First I tried: 1) Fixing x, and having a variable y, 2) Fixing y, and having a variable x, 3) Approaching 0 when $y = x$ and all concluded to $0$. So I thought the limit probably is $0$. Using the definition of the limits, I have to show that $\forall \epsilon > 0, \exists \delta > 0: || x – x_0 || < \delta \Rightarrow ||f(x) – \ell || < \epsilon, x_0=(0,0), \ell = 0$.
From that: $ 0 < || (x,y) – (0,0)|| < \delta \Rightarrow x^2 + y^2 < \delta ^2 \Rightarrow x^2 < \delta ^2 \Rightarrow |x| < \delta$. Now as we know that $|\sin(x)| \leq x$:
$${|\frac{\sin(xy)}{y}}| \leq |\frac{xy}{y}| = |x|$$
and thus if we choose $\delta = \epsilon$ we get:
$${|\frac{\sin(xy)}{y}} – 0| \leq |\frac{xy}{y}| = |x| < \delta = \epsilon$$ which is what we want to show, and thus $\lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{y} = 0$, which when graphed I think shows that really this is the case. Is what I have done correct? Could I make it better? Thanks
Best Answer
It looks correct to me. But you could simply say that$$\lim_{z\to 0}\frac{\sin z}z=1\implies\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{xy}=1$$and that therefore\begin{align}\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}y&=\lim_{(x,y)\to(0,0)}x\frac{\sin(xy)}{xy}\\&=\left(\lim_{(x,y)\to(0,0)}x\right)\left(\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{xy}\right)\\&=0\times1\\&=0.\end{align}Actually, since $\frac{\sin(xy)}{xy}$ is undefined if $x=0$ or $y=0$, this only proves that$$\lim_{\substack{(x,y)\to(0,0)\\ xy\ne0}}\frac{\sin(xy)}{xy}=0.$$But, since $\frac{\sin(xy)}y=0$ when $x=0$ and it is undefined when $y=0$, that's not a problem.