<s>[INST] Evaluate $\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$ How should I determine the following limit?
$\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$
 [/INST] 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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