proof of $\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}$ for large $n$ I'm trying to conclude this question in which I just need to prove that:
$$0<\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}$$
and then I'll have that, for large $n$:
$$\begin{align}
\left\lvert\cos\left(\sqrt{x^2+\frac{1}{n}}\right)-\cos(x)\right\rvert &= \left\lvert2\sin\frac{\sqrt{x^2+\frac{1}{n}}-x}{2}\sin\frac{\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}}{2}\right\rvert\\
&\leq
2\sin\frac{\sqrt{x^2+\frac{1}{n}}-x}{2}\sin\frac{1}{2\sqrt{n}}\xrightarrow[n\to\infty]{} 0
\end{align}$$
because $\sin\frac{1}{2\sqrt{n}}\to 0$ and $\sin\frac{\sqrt{x^2+\frac{1}{n}}-x}{2}$ is bounded.
I liked the answers given in the question but I really need to solve it this way, I'm trying to investigate the inequation. I know that, the worst case is for $n$ so large that $\frac{1}{n}$ in the LHS of the inequality goes to $0$, and the best case is when $n=1$ and therefore we have:
$$\sqrt{x^2}+x<\sqrt{x^2+\frac{1}{n}}+x<\sqrt{x^2+1}+x\le\sqrt{2}+1$$ 
*for $0\le x\le 1$
but it doesn't help, I think.

Cross-multiply, square, subtract ${1\over n}$ from both sides to get
$$
0\le 2x^2+2x\sqrt{x^2+{1\over n}}
$$
Assuming $x\in[0,1]$ the above identity is trivial as each term on the RHS is non-negative. Now backtrack and get your desired identity.
