show that $f(x)=\sin x$ is continuous

Question

how do you show that $f(x)=\sin x$ is continuous at any point of its domain?

I was thinking about taking the limit and show that it is equal to $0$ such as

$$\lim\limits_{x \rightarrow x_0} (f(x) -f(x_0)) =0 $$

But, I can't figure out how to proceed

much appreciated

What is your definition of $\sin x$? – Alex Provost Jul 13 '17 at 18:08 — Alex Provost, Jul 13 '17 at 18:08

Dr. Sonnhard Graubner · Answer 1 · 2017-07-13T17:55:58.697

3

HINT: use that $$\sin(x)-\sin(x_0)=2 \sin \left(\frac{x}{2}-\frac{x_0}{2}\right) \cos \left(\frac{x}{2}+\frac{x_0}{2}\right)$$ and write the right Hand side in the form $$(x-x_0)\cdot \frac{\sin\left(\frac{x-x_0}{2}\right)}{\frac{x-x_0}{2}}\cdot\cos\left(\frac{x+x_0}{2}\right)$$

edited Jul 13 '17 at 17:55

answered Jul 13 '17 at 17:36

Dr. Sonnhard Graubner

95,283

Right, but this just shows continuity at $x=0$ implies global continuity. Continuity at $0$ is true since $\lim_{x \to 0} \frac{\sin x}{x} = 1$, which has a geometric proof. – MathematicsStudent1122 Jul 13 '17 at 17:40
but $$x=x_0$$ isn't Zero in the General case – Dr. Sonnhard Graubner Jul 13 '17 at 17:42
Is there a way to factor out $(x - x_0)$ ? – gegu Jul 13 '17 at 17:51
Simply multiply and divide by it... – Aditya Jul 13 '17 at 17:57

score 1 · Answer 2 · answered Jul 14 '17 at 03:33

1

If you define $\sin(x) = \displaystyle \lim_{N \to \infty}\sum_{k=1}^N \frac{(-1)^{n+1}x^{2n-1}}{(2n-1)!}$ you have that it's the uniform limit of continuous functions (on compact intervals, but around any point we can take a such a region) and thus continuous.

answered Jul 14 '17 at 03:33

David Bowman

5,572

thx. any chance you could look at another question? I reedited it but no adequate answer. https://math.stackexchange.com/questions/2358035/find-all-derivatives-of-fx-xn-1-x-where-n-is-a-positive-integer – gegu Jul 14 '17 at 03:46

score 1 · Answer 3 · answered Jul 14 '17 at 04:40

$$\sin (x+d)-\sin x=(\sin x \cos d+\cos x \sin d)-\sin x=$$ $$=(\sin x)(-1+\cos d)+\cos x \sin d=-2\sin x \sin^2(d/2)+\cos x \sin d.$$ So it suffices to prove that $\lim_{d\to 0}\sin d=0,$ because that implies $\lim_{x\to 0}2\sin^2(d/2) =0$ and hence that $\lim_{d\to 0}\sin (x+d)-\sin x=0.$

The geometric definition of $\sin d$, for real $d$, immediately implies that $|\sin d|\leq |d|$ for all real $d.$

show that $f(x)=\sin x$ is continuous

3 Answers3