Evaluate $\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$ How should I determine the following limit?
$\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$
 A: HINT
Use $\cos x - \cos y = -2 \sin(\frac {(x - y)} 2 ) \sin(\frac {(x + y)} 2 )$
Then $\lim _{x\to \infty }\sin(\frac {(\sqrt x - \sqrt {x-1})} 2 )=\lim _{x\to \infty }\sin(\frac 1 {2(\sqrt x + \sqrt {x-1})})=0$
Because $\sin(\frac {(\sqrt x + \sqrt {x-1})} 2) $ is bounded, it follows the limit is zero
A: The difference between $\sqrt{x}$ and $\sqrt{x-1}$ becomes arbitrary small as $x$ goes to infinity. Since $\cos$ is a uniformly continuous function, the difference between the cosine of two arbitrary close numbers is 0. So the limit is 0.
Uniform continuity is a stronger form of continuity. Having a bounded derivative is a sufficient condition for a function to be uniformly continuous.
This method requires some extra work to make a rigid proof, but it is a general method that works in many cases and it complies with intuition.
A: $$\cos\sqrt{x}-\cos\sqrt{x-1}=2\sin\frac{\sqrt{x}+\sqrt{x-1}}{2}\sin\frac{\sqrt{x-1}-\sqrt{x}}{2}=$$
$$=-2\sin\frac{\sqrt{x}+\sqrt{x-1}}{2}\sin\frac{1}{2(\sqrt{x-1}+\sqrt{x})}\rightarrow0$$
A: $\displaystyle  \lim_{x\rightarrow \infty}\bigg|\int^{\sqrt{x}}_{\sqrt{x-1}}\sin xdx\bigg| \leq \lim_{x\rightarrow \infty}\int^{\sqrt{x}}_{\sqrt{x-1}}1dx =  \lim_{x\rightarrow \infty}\left(\sqrt{x}-\sqrt{x-1}\right) =  \lim_{x\rightarrow \infty}\frac{1}{\left(\sqrt{x}+\sqrt{x-1}\right)}=0$
A: Direct approach:
Let  $f(x) := \cos\sqrt{x} - \cos\sqrt{x - 1}$.
By the mean value theorem, for two values $a, b \in \mathbb R$, there exists $t\in\mathbb [a, b]$ such that $\cos(b) - \cos(a) = \cos'(t)(b - a)$. Thus,$$
|f(x)|=|\cos\sqrt{x} - \cos\sqrt{x - 1}| \\
=|\cos'(t)(\sqrt{x} - \sqrt{x - 1})| \\
=|-\sin(t)|\cdot|\sqrt{x} - \sqrt{x - 1}|\\
\le|\sqrt{x} - \sqrt{x - 1}| = \sqrt{x} - \sqrt{x - 1} =: g(x)$$
for all $x>1$.

Assertion: $\lim_{x\rightarrow\infty} g(x) = 0$.
Proof: Let $\epsilon > 0$. Then
$$g(x) \le \epsilon\\
\Leftrightarrow\sqrt{x} \leq \epsilon + \sqrt{x - 1}\\
\Leftrightarrow x \leq \epsilon^2 + 2\sqrt{x - 1} + x -1\\
\Leftrightarrow x \geq 1+(1-\epsilon^2)^2/4 =: x_0$$
Thus, for all $x> x_0$, we have $|g(x)| = g(x) \le \epsilon$.

Together with $|f(x)|\le g(x)$, this yields $\lim_{x\rightarrow\infty} f(x) = 0$.
