Uniform Convergence of $f_n(x)= \sin(x+\frac 1n)$

Question

Prove that $\sin(x+(1/n)) \rightarrow \sin x$ uniformly on $R$ using Mean Value Theorem.

I'm not sure how to go about this. Because the problem asks to use MVT, I tried taking the derivative of the function in order to find the supremum and use that to show that the supremum was less than some epsilon to prove uniform convergence, but I am having a hard time.

Answer 1

For all $x\in \mathbb R$,

$$\sin\left(x +\frac 1n \right) - \sin x = \sin' (y)\left( x +\frac 1n -x\right)$$

for some $y\in (x, x+\frac 1n)$ by the MVT. Note that the right hand side is always smaller than $\frac{1}{n}$.

Answer 2

Since $|\sin'(x)| \le 1$ for all $x$, we have $| \sin(x+ {1\over n})-\sin x| \le {1 \over n}$ for all $x$.

Answer 3

For all $n \geq 1$, by mean-value theorem we have $$ \Big| \sin \Big( x + \frac{1}{n} \Big) - \sin x \Big| \leq \Big( x + \frac{1}{n} - x \Big) \sup_{x < t < x + \frac{1}{n}}| \cos t| \leq \frac{1}{n} $$ for all $x \in \mathbb{R}$; if $\varepsilon > 0$, then $n \geq \Big\lceil \frac{1}{\varepsilon} \Big\rceil + 1$ only if $\frac{1}{n} < \varepsilon$; we have thus proved this: for every $\varepsilon > 0$, we have $n \geq \Big\lceil \frac{1}{\varepsilon} \Big\rceil + 1$ only if $$ \Big| \sin \Big( x + \frac{1}{n} \Big) - \sin x \Big| < \varepsilon $$ for all $x \in \mathbb{R}$