Show that $f(x,y)=\begin{cases} \frac{\phi(x) - \phi(y)}{x-y} & x \not = y\\ \phi'(x) & x=y \end{cases}$ is continuous

Question

Let $\phi \in C^1(\Bbb R; \Bbb R) $ and $f: \Bbb R^2 \to \Bbb R;$

$f(x,y)=\begin{cases} \frac{\phi(x) - \phi(y)}{x-y} & x \not = y\\ \phi'(x) & x=y \end{cases}$

Show that $f$ is continuous.

My attempt:

Clearly $f$ is continous on $\Bbb R^2 \backslash \{(x,y) \in \Bbb R^2:x=y\}$ since $\phi$ is continuous and the expression $\frac{\phi(x) - \phi(y)}{x-y}$ is well defined for $x \not = y$.

Continuitiy at $x = y$ :

Let $\mathbf q_n = (x_n, y_n)$ be a sequence such that $\mathbf q_n \to (c,c)$ for a $c \in \Bbb R^2$.

I need to show that

$$f(\lim_{n\to \infty}\mathbf q_n) = \lim_{n\to \infty} f(\mathbf q_n)$$

Using the definition of $f$ we get

$$f(\lim_{n\to \infty}\mathbf q_n) = f(c,c) = \phi'(c) = \lim_{x\to c} \frac{\phi(x)-\phi(c)}{x-c} $$ and

$$ \lim_{n\to \infty} f(\mathbf q_n) = \lim_{n\to \infty} \frac{\phi(x_n)-\phi(y_n)}{x_n-y_n}$$

which leads to equality immediately if $\mathbf q_n = (x_n, c)$.

Here I am not sure how to continue, I tried to use L'Hôpital on the last expression and got $$\lim_{n\to \infty} \frac{\phi(x_n)-\phi(y_n)}{x_n-y_n} = \lim_{n\to \infty} \frac{\frac{d}{dn}(\frac{d}{d \phi}(x_n \cdot \phi(x_n) - y_n \cdot \phi(y_n))}{\frac{d}{dn}(x_n - y_n)}$$

But that does not seem to go anywhere.

How should I proceed to prove continuity?

Edit:

Using the suggestion from @J.G:

Let $h_n = y_n - x_n$. Then

$$\lim_{n\to \infty} \frac{\phi(x_n)-\phi(y_n)}{x_n-y_n} = \lim_{n\to \infty} \frac{\phi(x_n)-\phi(x_n + h_n)}{-h_n} = \lim_{n\to \infty} \frac{\phi(x_n + h_n)-\phi(x_n)}{h_n}$$

Can I just set this expression equal to $$ \lim_{h\to 0} \frac{\phi(c + h)-\phi(c)}{h} ?$$

I feel like I am almost there.

Answer 1

Hint:$$\lim_{y\to x}f(x,y)=\lim_{y\to x}\frac{\phi(x) - \phi(y)}{x-y}=\phi'(x)$$

Answer 2

Maybe this argument is correct:

Be $\epsilon>0$ given.

Take $p:=(a,a)$ with $a\in \mathbb R$. Taking a sufficiently small positive $\delta$, $\phi$ being $C^1$ guarantees that $$|(x,y)-p|=\max\left \{|x-a|,|y-a| \right \}<\delta$$ implies $|\phi'(x)-\phi'(a)|<\epsilon$. If $x=y$, there's nothing to prove; otherwise, apply the Mean Value Theorem to the function $\phi$ considering $x,y\in(a-\delta,a+\delta)$, arriving at $|f(x,y)-\phi'(a)|<\epsilon$.