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How should I determine the following limit?

$\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$

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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

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.

$$\cos\sqrt{x}-\cos\sqrt{x-1}=2\sin\frac{\sqrt{x}+\sqrt{x-1}}{2}\sin\frac{\sqrt{x-1}-\sqrt{x}}{2}=$$

$$=-2\sin\frac{\sqrt{x}+\sqrt{x-1}}{2}\sin\frac{1}{2(\sqrt{x-1}+\sqrt{x})}\rightarrow0$$

$\displaystyle \lim_{x\rightarrow \infty}\bigg|\int^{\sqrt{x}}_{\sqrt{x-1}}\sin xdx\bigg| \leq \lim_{x\rightarrow \infty}\int^{\sqrt{x}}_{\sqrt{x-1}}1dx = \lim_{x\rightarrow \infty}\left(\sqrt{x}-\sqrt{x-1}\right) = \lim_{x\rightarrow \infty}\frac{1}{\left(\sqrt{x}+\sqrt{x-1}\right)}=0$

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 $\lim_{x\rightarrow\infty} f(x) = 0$.

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