If $f$ and $\phi$ are two functions from $\mathbb{R}$ to $\mathbb{R}$. $\varphi$ is the function from $\mathbb{R^2}$ to $\mathbb{R}$ defined by : $$\varphi(x,y)=\left\lbrace\begin{matrix}\frac{\phi(x)-\phi(y)}{x-y} \ \ if \ \ x\neq y \\ f(x) \ \ if \ \ x=y\end{matrix}\right.$$
I want to find the optimal condition on $f$ and $\phi$ so that $\varphi$ is continuous at $(1,1)$.
Also, is there a sufficient and necessary condition on $f$ and $\phi$ such that $\varphi$ is continous at $(1,1)$?
Thank you.
Suppose that $\varphi$ is continuous at $(1,1)$. Then $$\lim_{x \to 1} \varphi(x,1) = \varphi(1,1)$$ which translates to $$\lim_{x \to 1} \frac{\phi(x) - \phi(1)}{x-1} = f(1).$$ Thus it is necessary that $\phi$ is differentiable at $x=1$ and $\phi'(1) = f(1)$.
This should also be sufficient - the argument that follows isn't particularly rigorous but can be made so.
Suppose that $\phi$ is differentiable at $1$ and that $\phi'(1) = 1$. Suppose that $x,y$ are both near $1$. If $x \not= y$ and both $x \not=1$ and $y \not= 1$ then \begin{align*}\varphi(x,y) &= \frac{\phi(x) - \phi(y)}{x-y} \\ &= \frac{\phi(x) - \phi(1)}{x-y} + \frac{\phi(1) - \phi(y)}{x-y} \\ &= \frac{\phi(x) - \phi(1)}{x-1} \frac{x-1}{x-y} + \frac{\phi(y) - \phi(1)}{y-1} \frac{1-y}{x-y} \\ &\approx f(1) \frac{x-1}{x-y} + f(1)\frac{1-y}{x-y} \\ &= f(1) \\ &= \varphi(1,1). \end{align*}