# How to prove a function is always continuous? [closed]

For example, how could I prove that $\sin x$ is continuous for ]$-\infty,\infty$[ ?

## closed as unclear what you're asking by Mark McClure, JMP, user296602, Claude Leibovici, user91500Aug 5 '16 at 6:34

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• Just show that it is continuous for any $x\in\mathbb{R}$. – Marra Aug 4 '16 at 17:34
• See here for the $\sin x$ case in particular: cut-the-knot.org/do_you_know/CosineIsContinuous.shtml – smcc Aug 4 '16 at 17:35
• What is your definition for $\sin(x)$? – Andres Mejia Aug 4 '16 at 17:35
• For arbitrary function, you can only use the definition, there is no other way. – Michael Freimann Aug 4 '16 at 17:37
• As others said, it depends on how you define $\sin$. One definition is that is the unique function satisfying the differential equation $f''(x) = -f(x)$ subject to the constraints $f(0) = 0$ and $f'(0) = 1$. From this definition, continuity is immediate, as any differentiable function is continuous. (Of course, the hard part is showing that a solution to this differential equation exists!) – Bungo Aug 4 '16 at 17:47

Let $x \in \mathbb{R}$.

Take $$\sin(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}.$$

Then each of the partial sums are continuous. See here for a proof that the power series expansion converges uniformly (on the interval of convergence.) A sequence of continuous converges uniformly to $\sin(x)$, then $\sin(x)$ is continuous at $x$.

• Uniformly on compact intervals, right? – Carsten S Aug 5 '16 at 0:25
• I don't believe that link contains a proof that sine is analytic, which is a necessary precondition if you're going to take its Maclaurin series at some arbitrary $x$ without proving that $x$ lies in the interval of convergence. Furthermore, analyticity is a much stronger condition than continuity. – Kevin Aug 5 '16 at 2:56
• I am taking the power series to be the definition of the $\sin$ function. – Andres Mejia Aug 5 '16 at 3:01
• That is not the usual definition of the sine function. The sine function is defined in terms of the geometry of a right triangle, at least in real analysis. – Kevin Aug 5 '16 at 5:36
• I'm well aware that its not usual. That doesnt preclude it as a potential definition. The def. Wasnt specified, so I was having fun. There are other answers that use a definition to your liking. lighten up, there are different ways to solve a problem. – Andres Mejia Aug 5 '16 at 14:09

Holds more than that, $\sin{x}$ is uniformly continuous by $\varepsilon-\delta$ definition.

It is enough to choose $\delta=\varepsilon$ and the implication from the definition holds, since $$|\sin{x}-\sin{a}|=|2\sin{\frac{x-a}{2}}\cos{\frac{x+a}{2}}|<2|\frac{x-a}{2}||1|=|x-a|,$$ where we used known inequalities $\sin{t}<t$ and $\cos{t}<1$.

• Even better, the same inequality shows that $\sin x$ is Lipschitz continuous. – Bungo Aug 4 '16 at 17:50
• Good point! +1. – Andres Mejia Aug 4 '16 at 17:55

That depends on your definition of sin(x). Is it only intuitive or series or Euler's formula?

For the latest is the proof imho the simplest, using the arithmetic of continuous functions and the fact that the exponential function is a continuous function.

The proof can be worked out even for the definition via the power series, see the results concerning analytic functions.

You could show that $\sin(x)$ belongs to $C^{\infty}(\mathbb{R})$. Indeed the function is infinitely times differentiable. Since a function is which is differentiable must be continuous, the original function is continuous. This proof idea assumes you can take the fact that the derivative of $sin$ is $cos$ for granted. Should you not have that at your disposal, using the power series expansion argument already answered here would be sufficient.