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Let $f$ be given as

$$ f(x,y) = \begin{cases} \dfrac{ \sin x - \sin y }{x-y}, & \text{if }\text{ $x \neq y $} \\ \cos x, & \text{if } x \text{ $=y$} \end{cases} $$

My claim is that the function is discontinuous along the diagonal. But how can I show this?

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  • $\begingroup$ Try fixing $x$ and letting $y\to x$, then the other way round, fixing $y$ and letting $x\to y$. $\endgroup$ Feb 19, 2014 at 8:05

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You can write your function as $$f(x,y)=\frac{ \sin x - \sin y }{x-y}=\frac{2 \sin \left(\frac{x-y}{2}\right) \cos \left(\frac{x+y}{2}\right)}{x-y}$$ from which it is easy to see what happens along the diagonal (or close to it).

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  • $\begingroup$ can you explain a little bit more? I dont see how this helps... $\endgroup$
    – user124140
    Feb 19, 2014 at 12:10
  • $\begingroup$ @Learner. Use the fact that $\frac{\sin (a)}{a}\to 1$ when $a$ is small and replace $a$ by $\frac{x-y}{2}$ $\endgroup$ Feb 19, 2014 at 12:46

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