Is $\frac{\sin(xy)}{xy}$ continuous?

Question

Let $h: \mathbb{R}^2 $ -> $\mathbb{R}$

I can replace $h(x, 0) = x$, so that $x=a≠0$. Then the function essentially becomes $h(y) = \frac{\sin (ay)}{ay}$. To figure out the limit for "$h(0)$", we can substitute $z = ay$: $$ \lim_{y \to 0}f(a, y) = \lim_{z \to 0}\frac{\sin z}{z} = ? $$ For $x = 0$, we have that $h(0, y) = 0$ for non-zero $y$, so $h(0, 0) = 0$ is a natural extension at the origin as well.

Is that correct, or is that wrong? Can one do this differently?

Answer 1

Exactly the same reason with $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ shows that $$\lim\limits_{(x,y)\to(0,0)}\frac{\sin xy}{xy}=1$$ with the common formula $$\cos xy\leq\dfrac{\sin xy}{xy}\leq1$$ we can conclude the lmit is $1$ as the limit in one variable shows.

Answer 2

The best way to do this is to write $h$ as the composition of the maps $(x,y) \to xy$ and the map $g:\mathbb R \to \mathbb R$ defined by $g(x)=\frac {sin (x)} x$ if $x \neq 0$ and $g(0)=1$.

Answer 3

1. Show that for any non-zero x, as y approaches 0 from both sides ($0^{+}$ and $0^{-}$), h approaches 1.
2. Show that for any non-zero y, as x approaches 0 from both sides ($0^{+}$ and $0^{-}$), h approaches 1.

Since approaching (0, 0) from any direction is a linear combination of just x approach 0 and just y approach 0, steps 1 and 2 show that h approaches 1 when both x and y approaches 0 from any direction. This argument is similar to a multivariable Taylor expansion of the function.

Answer 4

As already noticed your claim $h(x, 0) = x $ is uncorrect since we obtain $$xy=0\cdot y=0 \implies \frac{\sin 0}{0}=1$$

to take the limit simply observe that $$(x,y) \to (0,0)\implies xy\to 0$$

then indicating $xy=z \to 0$ the limit becomes

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