Let $f:\mathbb R\to \mathbb R$ be a $C^1$ function.

Then Prove or disprove that $g:\mathbb R^2\to \mathbb R$ define by $$g(x,y) = \begin{cases}\frac{f(x)-f(y)}{x-y}&\text{if $x\neq y$}\\ f'(x)&\text{if $x= y$}\end{cases}$$ is a continuous function.

Note that $C^1$ Continuity must be crucial since a blatant counter example given by the function $$f(x) = \begin{cases}x^2\sin \frac{1}{x}&\text{if $x\neq 0$}\\ 0&\text{if $x= 0$}\end{cases}$$ which differentiable on $\mathbb R$ but not $C^1$ because the derivative is not continuous at $x=0.$


first we see that g is continuous at $(x,y)$ such that $x\neq y$ now let $x_{0} = y_{0}$ and let $ (x,y)$ converge to $(x_{0},y_{0})$ we have $\frac{f(y)-f(x)}{y-x}= {f}'(c) $ such that $ x\leq {c} \leq y $ . and the conclusion follow since $ f$ is $C^{1}$ continuous

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