# Prove that $\displaystyle{\lim_{x \to a}} \sin x = \sin a$ using epsilon-delta definition [closed]

I've been struggling with this question. I don't know how to get rid of the sine function after getting to $$|\sin x - \sin a| \leq 2 \left| \sin \dfrac{x - a}{2} \right|$$ Half angle formula isn't really useful imo, not sure what to do from here.

• $|\sin x|\leqslant |x|$ for all $x\in\mathbb{R}$.
– RRL
Jan 25, 2021 at 3:11

Thinking about an identity was a good idea. However, use $$\sin(\alpha)-\sin(\beta)=2\cos(\frac{\alpha+\beta}{2})\sin(\frac{\alpha-\beta}{2})$$.

We want to chek that $$\forall \epsilon>0$$, $$\exists$$ $$\delta>0$$ such that:

$$|x-c|<\delta \implies |\sin(x)-\sin(c)|<\epsilon$$.

Let $$\epsilon=\delta$$, then $$|\sin(x)-\sin(c)|=2\cos(\frac{x+c}{2})\sin(\frac{x-c}{2})\leq2|\frac{x-c}{2}|=|x-c|<\delta=\epsilon$$.

We can justify this last step with the following:

• $$|\cos(x)|\leq 1 ~\forall x\in\mathbb{R}$$
• $$|\sin(x)|<|x| ~\forall x\in\mathbb{R}$$

Use the mean value theorem. That is, $$|\sin(x) - \sin(a)| \leq |\cos(\psi)||x-a|$$ for some $$\psi \in (x, a)$$. I will let you finish the proof.

• this question is essentially asking to show that sine is continuous. are you not assuming that it is continuous by assuming that it is differentiable in the mean value theorem? Jan 25, 2021 at 4:08
• @C Squared you are correct I overlooked this. Jan 25, 2021 at 4:12