n such that $|\sin (\sqrt{n+1})-\sin \sqrt n|< \lambda$ 
I broke $|\sin (\sqrt{n+1})-\sin \sqrt n|$ as $|2cos(\frac{\sqrt{n+1}+\sqrt n}{2})sin(\frac{\sqrt{n+1}-\sqrt n}{2})|$.I am facing trouble in proving that it is less than some number for some n.Please help me in his regard.Thanks.
 A: I think a good idea would be to use the Mean Value theorem. Hence, for all positive integers $n$, you must have : 
\begin{equation*}
|\sin(\sqrt{n+1}) - \sin(\sqrt{n})| \leq | \sqrt{n+1} - \sqrt{n} | = \frac{1}{\sqrt{n+1} + \sqrt{n}} \underset{n \to \infty}{\longrightarrow} 0 
\end{equation*}
Now it is easy to see that the good answer must be $C$. 
A: You're on the right track.  Note that we have
$$\begin{align}
\left|\sin(\sqrt{n+1})-\sin(\sqrt n)\right|&=\left|2\cos\left(\frac{\sqrt{n+1}+\sqrt n}{2}\right)\sin\left(\frac{\sqrt{n+1}-\sqrt n}{2}\right)\right|\\\\
&\le 2\left|\sin\left(\frac{1}{2(\sqrt{n+1}+\sqrt n)}\right)\right| \tag 1\\\\
&\le \frac{1}{\sqrt{n+1}+\sqrt{n}} \tag 2\\\\
&\le \frac{1}{1+\sqrt2}
\end{align}$$
where in going from $(1)$ to $(2)$ we simply used the fact that $\sin(x)\le x$ for $x\ge 0$.
A: Note that for $0< x < \pi/2$, $\sin x < x$.
Also, for all $n > 0$,  we have:
$$
\frac{1}{4n} + 1 + n > n+1 \\
\left( \frac{1}{2\sqrt{n}}+\sqrt{n}\right)^2 > n+1 > 0 \\
\frac{1}{2\sqrt{n}} +\sqrt{n}> \sqrt{n+1} \\
\sqrt{n+1}-\sqrt{n}<\frac{1}{2\sqrt{n}}
$$
Also it is obvious that 
$$
\left|\cos \left( \frac{\sqrt{n+1}+\sqrt{n}}{2}\right)\right| \leq 1
$$
So for any $\lambda > 0$ and sufficiently large $n$,
$$
|\sin \sqrt{n+1}-\sin\sqrt{n} |< \lambda$$
Thus the answer is C.
