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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!

Chris Tang
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3 Answers3

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$\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

Jo Jomax
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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':

$\quad \displaystyle \lim_{x\to 0^{+}}\frac{\sin x}{x^2}=\lim_{x\to 0^{+}}(\frac{\sin x}{x})(\frac{1}{x})= (\lim_{x\to 0^{+}}\frac{\sin x}{x})(\lim_{x\to 0^{+}}\frac{1}{x})= +1 \times +\infty = +\infty$

$\quad \displaystyle \lim_{x\to 0^{-}}\frac{\sin x}{x^2}=\lim_{x\to 0^{-}}(\frac{\sin x}{x})(\frac{1}{x})= (\lim_{x\to 0^{-}}\frac{\sin x}{x})(\lim_{x\to 0^{-}}\frac{1}{x})= +1 \times -\infty = -\infty$

CopyPasteIt
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    Thank you, that helps; but is it legal to extend limits like this when one of them doesn't exist? (You may refer to Eevee Trainer's comment on the question.) – Chris Tang Feb 08 '20 at 14:27
  • If extended limits are in play then the limit of a product is the product of the limits but you have to be wary of the indeterminate form $0 \times \pm\infty$. – CopyPasteIt Feb 08 '20 at 15:05
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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.

PrincessEev
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