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.
 A: Well, as $x$ is positive, by monotonicity of the square root
$$\sqrt{x^2+\frac1n}+x \geq \sqrt{x^2+\frac1n}+\geq \sqrt{\frac1n}=\frac1{\sqrt{n}}$$
Thus
$$\frac{\frac{1}{n}}{\sqrt{x^2+\frac1n}+x}\leq \frac{\frac{1}{n}}{\sqrt{x^2+\frac1n}}\leq \frac{\frac{1}{n}}{\frac1{\sqrt{n}}}=\frac{1}{\sqrt{n}}$$
A: Outline: setting $\alpha=\frac{1}{\sqrt{n}}$, you want to prove
$$
\frac{1}{\sqrt{x^2+\alpha^2}+x} \leq \frac{1}{\alpha}, \qquad \forall x\geq 0\tag{1}
$$
Let us prove it for all $\alpha > 0$: this will then be equivalent to proving that for all $x\geq 0$
$$
\sqrt{x^2+\alpha^2}+x \geq \alpha, \qquad \forall x\geq 0
$$
that is, removing $x$ from each side and squaring, this will be equivalent* to showing that
$$
x^2+\alpha^2 \geq x^2+\alpha^2 - 2\alpha x, \qquad \forall x\geq 0.
$$
Now, why is that the case?

${}$* You may want to check that this is indeed equivalent, since we square... but yes, it is, as if the original RHS is negative then there is nothing to prove (the LHS is always positive).
A: 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.
