Based on this question, there is $\lim_{x\to 0}\frac{\sin x}{x}$; and how can I conclude $\lim_{x\to 0}\frac{\sin x}{x^2}$ doesn't exist from it, without using L'Hopital Rule?
I tried this way: $$\lim_{x\to 0}\frac{\sin x}{x^2}=\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{1}{x}=1\cdot\pm\infty$$ Yet I doubt myself because I know the limit law applies only when both limits exist, which really confuses me in many cases of using limit laws reversely.
I would be thankful for your help!
$\lim_{x\to 0^+} \frac{\sin x}{x^2}=\lim_{x\to 0^+} \frac{\sin x}{x}.\frac{1}{x}=+\infty $
$\lim_{x\to 0^-} \frac{\sin x}{x^2}=\lim_{x\to 0^-} \frac{\sin x}{x}.\frac{1}{x}=-\infty $
So there is no limit
Think about the following...
Let $b_n$ be a any bounded sequence and $c_n$ be any unbounded sequence.
Form the product sequence $a_n$ where $a_n = b_n c_n$
If $b_n = 0$ for all $n$ then $a_n$ converges.
Darn, and I wanted to state the following as true:
$\quad$ The sequence $a_n$ can't converge to any real number.
Could $a_n$ converge to $+\infty$ or $-\infty$?
If you are interested in extended limits from the left and right sides then this is the 'scoop':
$\quad \displaystyle \lim_{x\to 0^{+}}\frac{\sin x}{x^2}=\lim_{x\to 0^{+}}(\frac{\sin x}{x})(\frac{1}{x})= (\lim_{x\to 0^{+}}\frac{\sin x}{x})(\lim_{x\to 0^{+}}\frac{1}{x})= +1 \times +\infty = +\infty$
$\quad \displaystyle \lim_{x\to 0^{-}}\frac{\sin x}{x^2}=\lim_{x\to 0^{-}}(\frac{\sin x}{x})(\frac{1}{x})= (\lim_{x\to 0^{-}}\frac{\sin x}{x})(\lim_{x\to 0^{-}}\frac{1}{x})= +1 \times -\infty = -\infty$
It appears we may assume that the limit $\sin(x)/x \to 1$ (as $x \to 0$) is known. This establishes an asymptotic equivalence between $\sin(x)$ and $x$ as $x\to 0$. Therefore, we can swap these expressions in the limit, and therefore
$$\lim_{x \to 0} \frac{\sin(x)}{x^2} = \lim_{x \to 0} \frac{x}{x^2} = \lim_{x \to 0} \frac 1 x$$
This limit obviously does not exist.
