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

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?
1. Are there other ways to do it ?

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 $$1=\color{red}{4}L$$

• I hope my account won't be closed. If not, please delete this answer. Commented Jul 28, 2022 at 16:33
• hamam_ Abdallah. How did you get $1=4L$
– Gob
Commented Jul 28, 2022 at 16:49
• @Gob from the second equality with the red four. Commented Jul 28, 2022 at 17:12
• hamam_abdallah . But this only shows limit $\sin^2(x)$ does not exist.
– Gob
Commented Jul 28, 2022 at 17:18
• @Gob $\lim_{x\to\infty}\frac{x^2}{x^2+1}=1$ Commented Jul 28, 2022 at 17:46