Consider the following the function $f(x)=\frac{x^3\sin(x)}{x^2+1}$. What is the $\lim_{x\to \infty} \frac{x^3\sin(x)}{x^2+1}$?
My answer the limit doest not exist because $x_n=n\pi$ goes to $0$ as $n\to\infty$ and $f(x_n)=0$ for all $n$ so $f(x_n)$ goes to $0$ as $n\to \infty.$ Also, take $y_n=(2n+1)\frac{\pi}{2}$ goes to $\infty$ as $n\to \infty$ but $f(y_n)=\frac{y_n ^3}{y_n^2+1} (-1)^{n}$. So, $\lim_{n\to\infty} f(y_n)$ does not exist. Thus, $\lim_{x\to \infty} \frac{x^3\sin(x)}{x^2+1}$ does not exist.
Is that right? Is there another way to do it?
Yes, your idea is correct. But you want the $y_n = (4n+1)\dfrac{\pi}{2}$ instead. Then $f(y_n) \to \infty$, and $f(x_n) \to 0$ where $x_n = n\pi$. So you have $2$ different limits and hence no limit !