Let's do some calculus! (44)

this can be done by squeezing: $\lim_{x \to \infty} \dfrac{x^{101}}{e^x \arctan \left( \frac \pi3 + \sin x \right)} \geq \lim_{x \to \infty} \dfrac{x^{100}\ln(x)}{e^x \arctan \left( \frac \pi3 + \sin x \right)}\geq \lim_{x \to \infty} \dfrac{x^{100}}{e^x \arctan \left( \frac \pi3 + \sin x \right)}$ since $-1\leq\sin(x)\leq 1$ , $\arctan \left( \frac \pi3 -1 \right)\geq\arctan \left( \frac \pi3 + \sin x \right)\geq\arctan \left( \frac \pi3 + 1 \right)$ since arctangent is monotone in that range. but these are constants, so $0=\lim_{x \to \infty} \dfrac{x^{100}}{e^x \arctan \left( \frac \pi3 +1 \right)}\geq \lim_{x \to \infty} \dfrac{x^{100}}{e^x \arctan \left( \frac \pi3 + \sin x \right)}\geq \lim_{x \to \infty} \dfrac{x^{100}}{e^x \arctan \left( \frac \pi3 - 1 \right)}=0\to \lim_{x \to \infty} \dfrac{x^{100}}{e^x \arctan \left( \frac \pi3 + \sin x \right)}=0$ similarly $\lim_{x \to \infty} \dfrac{x^{101}}{e^x \arctan \left( \frac \pi3 + \sin x \right)}=0\to 0\geq\lim_{x \to \infty} \dfrac{x^{100}\ln(x)}{e^x \arctan \left( \frac \pi3 + \sin x \right)}\geq 0\\\lim_{x \to \infty} \dfrac{x^{100}\ln(x)}{e^x \arctan \left( \frac \pi3 + \sin x \right)}=\boxed{0}$

Regardless of the heavy stuff this problem has, look at the bottom, we have $e^x$ . Plugging $\infty \text {into}\ e^x$ yields an infinity denominator, thus the answer is 0 because $\frac{N}{\infty}=0$ Not a solution, just an idea that came to mind.