Problem : Find the limit of $$\lim_{x\to \infty}\frac{1}{x^2\sin x}=?$$
let $\lim_{x\to \infty}\frac{1}{x^2\sin x}=L$ Lo we have :
$$\forall \epsilon >0 \ \ \ \exists M>0 \ \ :x>M \to |f(x)-L|<\epsilon$$
Let $\epsilon =1$ so :
$$ \exists M_1>0 \ \ :x>M_1 \to |\frac{1}{x^2\sin x}-L|<1$$
now $x=\lfloor M_1\rfloor \pi $
Thus there is no limit. Is it correct?
Consider what happens at $x_n=2n\pi-(1/n^3)$ and $y_n=2n\pi+(1/n^3)$. It is easy to observe that $f(x_n) \to - \infty$ and $f(y_n) \to\infty$ as $n\to\infty$. So the function oscillates infinitely as $x\to\infty$. We have disregarded the points $x=n\pi$ which don't lie in the domain of the function.
If your textbook follows the simpler definition of limits where function must be defined in a deleted neighborhood of the point where limit is taken, then just saying that the function is undefined at $x=n\pi$ is sufficient to say that the limit doesn't exist. But as you go up in the mathematical education, you will see that such points which don't lie in the domain of the function are disregarded in the definition of limit and thus you need more work to show that the limit does not exist.
