Evaluating $\lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;)$ $$\lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;)$$
I know that due to continuity of sine I can split it in: 
$$\lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;) = \lim_{n \to \infty}\sin\sqrt{n+1} -\lim_{n \to \infty}\sin\sqrt{n}$$
I also know that sine has its values in $[-1;1]$, but I don't know how to combine these pieces of information. 
 A: No, you certainly cannot split it up that way. You're applying this fact:


Lemma. Suppose $c_n=a_n+b_n$. If $\lim a_n$ and $\lim b_n$ both exist then $\lim c_n=\lim a_n+\lim b_n$.


But you can't apply that here because $\lim_{n \to \infty}\sin(\sqrt n)$ does not exist.
You can find this limit by first applying the Mean Value Theorem...
A: Since$$\sqrt{n+1}=\sqrt{n}\cdot\sqrt{1+\frac1n}=\sqrt{n}\left(1+\frac{1}{2n}+O\left(\frac1n\right)\right)=\sqrt{n}+\frac{1}{2\sqrt{n}}+O\left(\frac{1}{\sqrt{n}}\right),$$calculus gives$$\sin\sqrt{n+1}-\sin\sqrt{n}\sim\frac{\cos\sqrt{n}}{2\sqrt{n}}.$$Since the cosine is bounded, this $\to0$ as $n\to\infty$.
A: write it just as $\sin c-\sin d=2.\cos \frac{c+d}{2}.\sin\frac{c-d}{2}$ now, substituting ,
here, $\cos \frac{\sqrt {(n+1)}+\sqrt n }{2}$.$\sin \frac{\sqrt {(n+1)}-\sqrt n }{2}$
now see that $\sin \frac{\sqrt {(n+1)}-\sqrt n }{2}\to 0$ as ${n\to\infty}$.
and note here that $|\cos \frac{\sqrt {(n+1)}+\sqrt n }{2}|\leq1$, so the result is $0$
A: Your line,
$$  \lim_{n \to \infty}(\sin\sqrt{n+1} - \sin\sqrt{n}\;) \\
\qquad \qquad = \lim_{n \to \infty}\sin\sqrt{n+1} -\lim_{n \to \infty}\sin\sqrt{n}  \text{,}  $$
is only valid if the two limits on the right exist.  Neither limit on the right exists, so this is an invalid operation.
Generally, you can make progress on this type of expression using the sum-to-product trigonometric identiites.
$$  \sin \sqrt{n+1} - \sin \sqrt{n} \\
\qquad = 2 \sin\left(\frac{\sqrt{n+1} - \sqrt{n}}{2}\right) \cos \left( \frac{\sqrt{n+1} + \sqrt{n}}{2} \right)  \text{.}  $$
In this particular case, we need a little more to see the limit.  We use conjugates to rewrite the arguments to be decreasing towards $0$.  \begin{align*}
&2 \sin\left(\frac{\sqrt{n+1} - \sqrt{n}}{2}\right) \cos \left( \frac{\sqrt{n+1} + \sqrt{n}}{2} \right)  \\
&\quad{}= 2 \sin\left(\frac{\sqrt{n+1} - \sqrt{n}}{2} \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} \right) \cos \left( \frac{\sqrt{n+1} + \sqrt{n}}{2} \cdot \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} \right)  \\
&\quad{}= 2 \sin\left(\frac{(n+1) - (n)}{2(\sqrt{n+1} + \sqrt{n})}  \right) \cos \left( \frac{(n+1) + (n)}{2(\sqrt{n+1} - \sqrt{n})}  \right)  \\
&\quad{}= 2 \sin\left(\frac{1}{2(\sqrt{n+1} + \sqrt{n})}  \right) \cos \left( \frac{2n+1}{2(\sqrt{n+1} - \sqrt{n})}  \right)  \text{.}
\end{align*}
Now we can use the squeeze theorem.  $-1 \leq \cos(\text{anything}) \leq 1$, so $$  -2 \sin\left(\frac{1}{2(\sqrt{n+1} + \sqrt{n})}  \right)  
\\{}\leq  2 \sin\left(\frac{1}{2(\sqrt{n+1} + \sqrt{n})}  \right) \cos \left( \frac{2n+1}{2(\sqrt{n+1} - \sqrt{n})} \right) 
\\{}\leq  2 \sin\left(\frac{1}{2(\sqrt{n+1} + \sqrt{n})}  \right)   \text{.}  $$
Now we study \begin{align*}
& \lim_{n \rightarrow \infty} 2 \sin\left(\frac{1}{2(\sqrt{n+1} + \sqrt{n})}  \right)  \\
&= 2 \sin\left( \lim_{n \rightarrow \infty} \frac{1}{2(\sqrt{n+1} + \sqrt{n})}  \right)  \\
&= 2 \sin 0  \\
&= 0  \text{.}
\end{align*}
Therefore, the desired limit is squeezed between $-0$ and $0$ as $n \rightarrow \infty$.  That is, 
$$  \lim_{n \rightarrow \infty} \left( \sin\sqrt{n+1} - \sin\sqrt{n} \right) = 0  \text{.}  $$
