I will write my problem and my approach in detail as I am not really comfortable with sequences of functions and analyzing their convergence, any feedback or improvement is appreciated. I am asked to investigate the convergence and type of convergence of the sequence $(\sin\frac{x}{n},\cos\frac{x}{n})_{n}$ on $\mathbb{R}$. I investigate the coordinate sequences separately:
For the first coordinate, $\forall x\in \mathbb{R} \left| \sin\frac{x}{n} \right| \le \left| x \right| \frac{1}{n} \longrightarrow 0$, so the sequence $(\sin\frac{x}{n})_{n}$ converges pointwisely to the zero function.
For the second, $\forall x\in \mathbb{R} \left| \cos\frac{x}{n} -1 \right| = \left| \cos\frac{x}{n} -\cos0 \right| \le \left| \frac{x}{n} - 0\right| = \left| x \right| \frac{1}{n} \longrightarrow 0$. Thus, the sequence $(\cos\frac{x}{n})_{n}$ converges pointwisely to the constant function $1$.
From these two, I concluded that the sequence $(\sin\frac{x}{n},\cos\frac{x}{n})_{n}$ converges pointwisely to the function $f(x)=(0,1)$.
However, this convergence is not uniform on $\mathbb{R}$ because one of the coordinate convergences is not uniform: $ \left| \left| \sin\frac{x}{n} - 0 \right| \right| _{\infty } = \underset{x\in \mathbb{R}}{\text{sup}} \left| \sin\frac{x}{n} \right| =1 \nrightarrow 0 $.
Is this line of arguments correct? When showing both pointwise convergence and uniform divergence I used the coordinate sequences, is this valid?