Looking for another way to show limit DNE without using sequence argument

Question

I asked a question similar to this yesterday see, Using a sequence argument to show that $\lim_{x\to \infty} \frac{x^3\sin(x)}{x^2+1}$ does not exist. I ask this again since I could not do it without a sequence argument.

Consider the following the functions $f(x)=\frac{x^2\sin^2(x)}{x^2+1}$ and $g(x)=\frac{x^3\sin^2(x)}{x^2+1}$

What are the $\lim_{x\to \infty} f(x)$ and $\lim_{x\to \infty} g(x).$ My answer does not exist for both functions. The argument would be similar so I will do just for $f(x).$

Now, $f(x)=\frac{x^2}{x^2+1} \frac{1-\cos(2x)}{2}$. Take $x_n=n\pi$, clearly $x_n\to \infty$ as $n\to\infty$ and $f(x_n)\to 0$ as $n\to\infty.$ Also, take $y_n=(2n+1)\frac{\pi}{2}$. So, $y_n\to \infty$ as $n\to\infty$ and $f(y_n)\to \frac{\pi^2}{4}$ as $n\to\infty.$ So, we have $f(x_n)\neq f(y_n)$. Hence, $\lim_{x\to \infty} f(x)$ does not exist. In case, $\lim_{x\to \infty} g(x)$, the argument would be quite similar. Perhaps my questions are

1. Is that right?

2. Are there other ways to do it ?

Answer 1

hint

Let us prove that $$\lim_{x\to+\infty}\sin^2(x)$$ does not exist.

Assume that this limit $=L(\in \Bbb R)$.

we have $$\cos^2(x)=\sin^2(x+\frac{\pi}{2}),$$ $$\sin^2(2x)=\color{red}{4}\sin^2(x)\cos^2(x),$$ and $$\sin^2(x)+\cos^2(x)=1$$

So, we will have $$2L=1$$ and $$1=\color{red}{4}L$$