I have the sequence $(f_n)$ of functions $f_n: [-1, 1] \rightarrow \mathbb{R}$, defined by $f_n(x) = \cos\left(\frac{x}{n}\right)$. I need to show that $(f_n)$ is Cauchy in the space $(C[-1, 1], d)$ where $d(f, g)= \max\{|f(x)-g(x)| : x \in [-1, 1]\}$. Also, find a function $f:[-1, 1] \to \mathbb{R}$ to which $f$ converges in $(C[-1, 1], d)$
I am not sure that I understand this question. Doesn't $\cos(\frac{x}{n})$ converge to $\cos(0) = 1$? What can you show otherwise?
A function could be just the constant function $f(x) = 1$
For this problem, you need to use $$ \cos x-\cos y=-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}. $$ So $$ d(f_{n+p},f_n) \le \frac{1}{n(n+p)}\le\frac{1}{n}. $$ Thus $\{f_n(x)\}$ is Cauchy.