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The question was implicitely already discussed in this Post.

However, the only answer which used an argument similar to my proofs isn't very detailed and I was wondering whether or not my proofs actually hold.

We want to show that the function $$f: \mathbb{R}^2 \rightarrow \mathbb{R}, (x,y) \mapsto \begin{cases} \frac{\sin(x-y)}{x-y} &x \neq y \\ 1 &x=y \end{cases} $$ Is continuous.

My proofs: We write $f = (g \circ \Delta) (x,y)$, where

$$(x,y) \overset{\Delta(x,y)}\mapsto x-y \qquad t \overset{g}{\mapsto} \begin{cases} \frac{\sin(t)}{t} & t \neq 0 \\ 1 &t=0 \end{cases} $$

First Proof: The function $\Delta(x,y)$ is continuous on $\mathbb{R}^2$ and so is $g$, with $g(0) = 1$. Therefore, $f$ is continuous as the composition of two continuous functions on valid domains ($\Delta(x,y) \in \mathbb{R}$) and for $x=y$ we get $f(x,x) = g(\Delta(x,x)) = g(0) = 1$.

Second Proof: Let $(x_n,y_n) \rightarrow (x,y), (x_n,y_n) \neq (x,y)$. We have: $$\lim_{n \rightarrow \infty} \Delta(x_n,y_n) = \lim_{n} x_n - y_n = \lim_n x_n - \lim_n y_n = x-y = \Delta(x,y)$$ and so $\Delta$ is continuous.

For any sequence $t_n \rightarrow t \in \mathbb{R}, t_n \neq t$ we have directly for $t\neq0$ that $$\lim_n g(t_n) = \lim_n\frac{\sin(t_n)}{t_n} = \frac{\sin(t)}{t} = g(t)$$

In the case where $\lim_n t_n = 0$, we find (for ex de l'hopital or see here) that $$\lim_n g(t_n) = 1 = g(0) = g(\lim_n t_n)$$ hence $g$ is continuous.

Now, $t_n := \Delta(x_n,y_n) \in \mathbb{R}$ is a real sequence with $\lim_n t_n = x-y := t \in \mathbb{R}$ and so we get: $$\lim_n f(x_n,y_n) = \lim_n g(x_n - y_n) = \lim_n g(t_n) = g(\lim_n t_n) = g(\lim_n \Delta(x_n,y_n)) = g(\Delta(\lim_n(x_n,y_n)) = g(\Delta(x,y)) = f(x,y)$$

Hence, when $(x_n,y_n) \rightarrow (x,x)$ like before, we have $\lim_n t_n = 0$ and so $f(x,x) = g(0) = 1$.

Pastudent
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    This really follows from the continuity of $$g(x)=\begin{cases}1&x=0\\frac{\sin x}x&x\neq0\end{cases}$$ Then $f(x,y)=g(x-y),$ and $(x,y)\mapsto x-y$ is continuous. You can do it your way, but it will be much harder to read and understand, and prone to hidden errors. – Thomas Andrews Jul 12 '23 at 18:26
  • @ThomasAndrews I agree, writing $(x,y) \mapsto x - y \mapsto g(x-y)$ is what I usually do. But when I start moving limit signs around, it can get confusing, which was the reason I went this less readable way. – Pastudent Jul 12 '23 at 18:31
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    $\lim$ is usually taken to have higher precedence than $-$ so $\lim_n x_n - y_n$ should read $\lim(x_n - y_n)$. In any case, you need to prove the continuity of $g$ at $0$. You've hinted at that in your second proof but just stated it in your first proof. – Rob Arthan Jul 12 '23 at 19:22

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