Evaluate $\lim_{x\to \infty} \cos (\sqrt {x+1})-\cos (\sqrt x)$

Solution 1:

By your identity you have $$=2\sin\frac{\sqrt{x+1}+\sqrt{x}}{2}\sin\frac{\sqrt{x}-\sqrt{x+1}}{2}$$

Now$$\frac{\sqrt{x}-\sqrt{x+1}}{2}\to 0$$ And thus $$\sin\frac{\sqrt{x}-\sqrt{x+1}}{2}\to 0$$ However $2\sin\frac{\sqrt{x+1}+\sqrt{x}}{2}$ is bounded and so the limit is zero.

Solution 2:

By the mean value theorem, for any $x$ there is a $c\in(\sqrt{x},\sqrt{x+1})$ such that:

$$\cos\sqrt{x+1}-\cos\sqrt{x} = -\sin(c)\left(\sqrt{x+1}-\sqrt{x}\right).$$

And so:

$$\left|\cos\sqrt{x+1}-\cos\sqrt{x}\right|\leq\sqrt{x+1}-\sqrt{x}=\frac{1}{\sqrt{x}+\sqrt{x+1}}\to 0$$

You don't need differentiability.

Claim:

If $f$ is uniformly continuous, and $\lim_{x\to\infty} (a(x)-b(x))=0$, you can show $f(a(x))-f(b(x))\to 0.$

Proof:

Given $\epsilon>0,$ we find $\delta$ such that if $|x-y|<\delta$ then $|f(x)-f(y)<\epsilon.$ (Definition of uniform continuity.)

Then we find $N$ such that when $x>N$, $|a(x)-b(x)|<\delta$.

So if $x>N$, $|a(x)-b(x)|<\delta$ so $|f(a(x))-f(b(x))|<\epsilon,$ and hence $$\lim_{x\to\infty}\left(f\left(a(x)\right)-f\left(b(x)\right)\right)=0.$$

In particular, any continuous and periodic function (like $\cos(x)$) on $\mathbb R$ is uniformly continuous.

And if $a(x)=\sqrt{x+1}$ and $b(x)=\sqrt{x},$ then $a(x)-b(x)=\frac{1}{\sqrt{x+1}+\sqrt{x}}\to 0.$

Solution 3:

Note that

\begin{align*} \lim_{x\to \infty} \sin\frac{\sqrt{x+1}-\sqrt{x}}{2}&=\lim_{x\to \infty} \sin\frac{1}{2(\sqrt{x+1}+\sqrt{x})}=0 \end{align*}

Since

\begin{align*} \left|\cos (\sqrt {x+1})-\cos (\sqrt x)\right|&=\left|-2\sin\frac{\sqrt{x+1}+\sqrt{x}}{2}\sin\frac{\sqrt{x+1}-\sqrt{x}}{2}\right|\\ &\le 2\left|\sin\frac{\sqrt{x+1}-\sqrt{x}}{2}\right|\\ &\to 0 \end{align*} as $x\to\infty$,

\begin{align*} \lim_{x\to \infty} \left[\cos (\sqrt {x+1})-\cos (\sqrt x)\right]&=0 \end{align*}