Prove the Limit

[myra_panama]

Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function so that $\left|f(x)-\sin\left(x^2\right)\right|\le\frac 14$ for any $x\in\mathbb{R}$ . Prove that there exists a sequence of real numbers $(x_n)_{n\ge 1}$ for which $\lim\limits_{n\to\infty}\, f^{\prime}\left(x_n\right)=+\infty .$

[Руст]

Let $y_n=\sqrt{\frac{\pi (4n-1)}{2}}, z_n=\sqrt{\frac{\pi (4n+1)}{2}}$ and $x_n\in [y_n,z_n]$, suth that $f'(x_n)=\frac{f(z_n)-f(y_n)}{z_n-y_n}\to \infty.$

[myra_panama]

Let $a_n = \sqrt{(2n-\frac{1}{2})\pi}$ and let $b_n = \sqrt{(2n+\frac{1}{2})\pi}$ for all $n \in \mathbb{N}$.

Note that the intervals $(a_n,b_n)$ are non-overlapping for all $n \in \mathbb{N}$. Also, $\sin(a^2_n) = -1$ and $\sin(b^2_n) = 1 $ for all $n \in \mathbb{N}. $
For any $n \in \mathbb{N}$, we have: $|f(a_n) - \sin(a^2_n)| \le \frac{1}{4} \leadsto |f(a_n) + 1| \le \frac{1}{4} \leadsto f(a_n) \le -\dfrac{3}{4} $
For any n $\in \mathbb{N}$, we have: $|f(b_n) - \sin(b^2_n)| \le \frac{1}{4} \leadsto |f(b_n) - 1| \le \frac{1}{4} \leadsto f(b_n) \ge \dfrac{3}{4}$

By the mean value theorem, there exists an $x_n \in (a_n , b_n)$ such that:

$f'(x_n) = \dfrac{f(b_n) - f(a_n)}{b_n - a_n} \ge \dfrac{\frac{3}{4} - (-\dfrac{3}{4})}{\sqrt{(2n+\frac{1}{2})\pi} - \sqrt{(2n-\frac{1}{2})\pi}} = \dfrac{3}{2\sqrt{\pi}}\left(\sqrt{2n+\dfrac{1}{2}} + \sqrt{2n-\dfrac{1}{2}}\right) $
Clearly, $f'(x_n)$ becomes arbitrarily large as $n \to \infty$. Thus, selecting an $x_n$ this way in every interval $(a_n,b_n)$ yields a sequence of real numbers $(x_{n})_{n\ge 1}$ such that $\lim\limits_{n\to\infty}\, f^{\prime}\left(x_{n}\right)=+\infty$ as desired

[Kallikanzarid]

You can't prove the limit, and it's FABULOUS!!!