I am aware that sums, products and quotients of continuous multivariable functions gives continuous multivariable functions. My query is whether knowing that some constituent single variable function means the overall multivariable function is continuous.

The example I am currently working on is proving that the function given by:

f(x,y)=(sin(x-y))/(x-y) where x=/=y and f(x,y) = 1 when x=y is continuous everywhere

I have expanded the sin(x-y) into a product of single variable trig functions.

Since f is then a combination of continuous single variable functions does that mean the multivariable function is continuous?

If no would anyone be able to give me a hint as to how to use either the epsilon-delta or sequential definition to show that f is continuous everywhere.

Thanks in advance

  • $\begingroup$ How did you expand $\sin(x-y)$ into a product? $\endgroup$
    – zhw.
    Oct 6, 2018 at 18:45
  • $\begingroup$ Sin(x-y)=sin(X)cos(y)-sin(y)cos(X) $\endgroup$
    – user601175
    Oct 7, 2018 at 8:26
  • $\begingroup$ That's not a product, it's a difference of products. $\endgroup$
    – zhw.
    Oct 7, 2018 at 15:41

1 Answer 1


The function $g(u) = (\sin u)/u, u\ne 0,$ $g(0)=1,$ is continuous on $\mathbb R.$ Your function $f(x,y)$ equals $g(x-y)$ for all $(x,y).$ Thus $f$ is the composition of continuous functions, hence is continuous.

  • $\begingroup$ Would I not still need to give a proof that the g(u) you have given is continuous? Or does it follow from g'(u)= sin(u) being continuous and g''(u)=u being continuous $\endgroup$
    – user601175
    Oct 7, 2018 at 8:28
  • $\begingroup$ $g$ is continuous everywhere except possibly at $0.$ But at $0$ we know $g(u)\to 1 = g(0).$ I'm not sure what you are doing with those derivatives. $\endgroup$
    – zhw.
    Oct 7, 2018 at 15:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .