Prove that $\lim\limits_{n\to\infty} \frac{\sin\left(\log n + n^5 + e^{n^2}\right)}{n} = 0$. Prove that $\lim\limits_{n\to\infty}x_n = 0$, where $x_n = \frac{\sin(\log n + n^5 + e^{n^2})}{n}$.
Demostración: Suppose $\epsilon > 0$. We first note the $\left|\sin(x)\right| \leq 1$ for $x\in\mathbb{R}$. Thus, $\left|\sin\left(\log n + n^5 + e^{n^2}\right)\right|\leq 1$. Pick $N > \frac{1}{\epsilon} \implies \frac{1}{N} < \epsilon$. For $n\geq N $, we have $\frac{1}{n} \leq \frac{1}{N} < \epsilon$. Second note that $\frac{1}{n} > -\epsilon $. Thus, $\left|\frac{1}{n}\right| < \epsilon$. Finally,
$$\left|\frac{1}{n}\cdot \sin\left(\log n + n^5 + e^{n^2}\right)\right| < \epsilon.\text{ QED}$$
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HINT
We can use
$$-\frac1n \le \frac{\sin\left(\log n + n^5 + e^{n^2}\right)}{n}\le \frac1n$$
