$f_n(x) = \cos \left( \sqrt{x^2 + \frac{1}{n} }\right)$ is uniformly convergent I need to prove that this sequence of functions:

$$f_n(x) = \cos \left( \sqrt{x^2 + \frac{1}{n} }\right)$$ 

converges uniformly on $[0,1]$.

This is a question regarding this one. There's an answer that suggests these hints:

$$\cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$$

$$\sqrt{x^2+\frac{1}{n}}-x=\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}\tag{1}$$

in order to prove that, $\forall \epsilon&gt;0, \exists n_0$ such that, for all $n\in \mathbb{N}$ we have:

$$n&gt;n_0 \implies \sup\left|\cos\left(\sqrt{x^2+\frac{1}{n}}\right)-\cos(x)\right|&lt;\epsilon$$
I did the following:

If I prove that such $n_0$ always exists, I can prove that this sequence converges uniformly (or not). By the hints he gave, we have:

$$\left|\cos\left(\sqrt{x^2+\frac{1}{n}}\right)-\cos(x)\right| = \left|2\sin\frac{\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}-x}}{2}\sin\frac{\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}}{2}\right|\le 2\sin\frac{1}{2\sqrt{n}}\sin\frac{1}{2\sqrt{n}}\to 0$$

*the last $\le$ symbol holds because $\sin$ is crescent from $0$ to $\frac{\pi}{2}$, but how $x\in [0,1]$ guarantee that the following will happen? Also, what if $x&gt;1$ or $x&lt;0$? Don't inequality $(1)$ holds for all $x$?

So since $\left|\cos\left(\sqrt{x^2+\frac{1}{n}}\right)-\cos(x)\right|$ is always greater than $0$ but we proved here it's $\le 0$, we have that $\sup$ of this sequence of absolute values is $0$. We haven't touched $n_0$ so far, neither $\epsilon$. Is it because the inequalities used hold for large $n$ and we always have $0&lt;\epsilon$?. Therefore, I proved that this sequence converges uniformly.

Extra: also, does $\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\le \frac{1}{\sqrt{n}}$ holds because:

$$n\left(\sqrt{x^2+\frac{1}{n}}+x\right)\ge \sqrt{n}\implies LHS^{-1}\le RHS^{-1}$$

?

I think that since $n&gt;\sqrt{n}$ for $n&gt;1$ (we're intersted in large values of $n$), we have that $n\cdot \mbox{something}&gt;\sqrt{n}$.
Here's a different idea: The mean value theorem gives $\cos b - \cos a = -\sin c\cdot(b-a),$ which implies $|\cos b - \cos a| \le |b-a|.$ Thus

$$|\cos \sqrt {x^2+1/n} - \cos x| \le |\sqrt {x^2+1/n} - x|.$$

So we need only show $\sqrt {x^2+1/n} \to x$ uniformly on $[0,1].$ For this it's good to know that if $a,b \ge 0,$ then $\sqrt {a+b} \le \sqrt a + \sqrt b.$ (Square both sides to see this.) Thus

$$ x &lt; \sqrt {x^2+1/n} \le x + 1/\sqrt n.$$

This implies $0 &lt; \sqrt {x^2+1/n} - x \le 1/\sqrt n,$ and we're done. (In fact we have shown uniform convergence on $ [0,\infty).$)
Here's a different idea: The mean value theorem gives $\cos b - \cos a = -\sin c\cdot(b-a),$ which implies $|\cos b - \cos a| \le |b-a|.$ Thus

$$|\cos \sqrt {x^2+1/n} - \cos x| \le |\sqrt {x^2+1/n} - x|.$$

So we need only show $\sqrt {x^2+1/n} \to x$ uniformly on $[0,1].$ For this it's good to know that if $a,b \ge 0,$ then $\sqrt {a+b} \le \sqrt a + \sqrt b.$ (Square both sides to see this.) Thus

$$ x &lt; \sqrt {x^2+1/n} \le x + 1/\sqrt n.$$

This implies $0 &lt; \sqrt {x^2+1/n} - x \le 1/\sqrt n,$ and we're done. (In fact we have shown uniform convergence on $ [0,\infty).$)