Given $f_n(x) := \cos{(x+\frac{1}{\sqrt{n}})}$. Does ${f_n(x)}_{n \geq 1}$ converge uniformly on $\mathbb R$?

Question

Given $f_n(x) := \cos{(x+\frac{1}{\sqrt{n}})}$. Does ${f_n(x)}_{n \geq 1}$ converge uniformly on $\mathbb R$ ?

$f_n(x)\to\cos{(x)}$ as $n\to \infty$ $$\left|f_n(x) -\cos{(x)} \right| = \left| \cos{(x+\frac{1}{\sqrt{n}})} - \cos{(x)} \right| $$

How do I get the inequality to appear?

Answer 1

Note that $$\left|\cos\left(x + \dfrac{1}{\sqrt{n}}\right) - \cos(x)\right| = 2\left|\sin\left(x + \dfrac{1}{\sqrt{n}}\right)\sin\left(\dfrac{1}{2\sqrt{n}}\right)\right| \le 2\left|\sin\left(\dfrac{1}{2\sqrt{n}}\right)\right|.$$

Also note that $$\lim_{n\to\infty}\sin\left(\dfrac{1}{2\sqrt{n}}\right)=0.$$

Thus, given any $\epsilon > 0$, you can find an $n_0 \in \Bbb N$ such that $$|f_n(x) - \cos(x)| < \epsilon$$ for all $n > n_0$ and $x \in \Bbb R$.

Answer 2

You may use the mean value theorem to get

$$ |\cos(x+n^{-1/2}) -\cos(x)|=|\sin(x+\theta_xn^{-1/2})|\frac{1}{\sqrt{n}}\leq\frac{1}{\sqrt{n}} $$ for some $0<\theta_x<1$. From this it is clear that $f_n$ uniformly converges to $f(x)=\cos(x)$