If $x_n = \cos{\sqrt{n + 1}} - \cos{\sqrt{n}}$, what is $\lim_{x \to \infty} x_n$?

Question

If $x_n = \cos{\sqrt{n + 1}} - \cos{\sqrt{n}}$, what is $\lim_{n \to \infty} x_n$?

I think it may be $0$, because $\sqrt{n + 1}$ and $\sqrt{n}$ are two very close angles. But I don't know how to prove it. If I use the formula for $\cos - \cos$, I get $-0 \cdot \infty$.

Can you help me, please? Thanks!

Answer 1

Well $$\cos{\sqrt{n + 1}} - \cos{\sqrt{n}}=2\sin\frac12(\sqrt{n+1}+\sqrt n)\color{red}{\sin\frac12(\sqrt n-\sqrt{n+1})}$$ and as $n\to\infty$ the term in red goes to zero.

Answer 2

$f(n)=\dfrac{\cos \sqrt{n+1}-\cos √n}{1}=$

$\sin (√t) \cdot (1/2)t^{-1/2}$, where $t \in (n,n+1)$.

$n \rightarrow \infty $ implies $t \rightarrow \infty$.

$\lim_{n \rightarrow \infty}f(n)=$

$ \lim_{t \rightarrow \infty} (1/2)( \sin √t) t^{-1/2}=0.$

Used : MVT

Answer 3

$$|\cos x-\cos y|\le|x-y|$$ (consider the points $(\cos x,\sin x)$ and $(\cos y,\sin y)$ on the unit circle). Therefore $$|x_n|\le\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\to0.$$