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I needed to show that $\lim_\limits{x\to 0} asin(1/x)$ doesn't exist when $a \neq 0$. This can be clearly done using $\epsilon-\delta$ method.

But in one of the proposed solutions following argument was used.

$\lim_\limits{x\to 0} asin(1/x)=a\lim_\limits{x\to 0} sin(1/x) \cdots\cdot (1)$

And since $\lim_\limits{x\to 0} sin(1/x)$ doesn't exist hence above limit also doesn't exist.

I have some reservations in the above step (1) which seems very obvious. Is it valid to separate constant as such when the limit doesn't exist?

Also, if it is valid then how to reason on above lines when $a=0$.

user512074
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    Yes this is a valid use of limit laws. For details see https://math.stackexchange.com/q/2971122/72031 – Paramanand Singh Jan 10 '21 at 12:29
  • The limit law does not allow you to factor out a $0$ and hence the same can't be done if $a=0$. But you can see that $a=0$ makes the function identically $0$ and computation of limit is trivial. – Paramanand Singh Jan 10 '21 at 12:30

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I believe the Theorem which you must use is this (and note that it has hypotheses as well as conclusions):

Theorem If $\lim_{x\to x_0}f(x)$ exists, and $\lambda\in\mathbb{R}$, then $\lim_{x\to x_0}\lambda f(x)$ exists and equals $\lambda\lim_{x\to x_0} f(x)$.

Now argue as follows when $a\not=0$.

Suppose that $\lim_{x\to 0} a \sin\frac{1}{x}$ exists. Then by the Theorem (with $\lambda=\frac{1}{a}$, $f(x)=a \sin\frac{1}{x}$) we get that $\lim_{x\to 0}\sin\frac{1}{x}$ exists. But we already know that this limit does not exist, so neither does $\lim_{x\to 0} a \sin\frac{1}{x}$.

When $a=0$ then $a\sin\frac{1}{x}=0$, and clearly $\lim_{x\to 0}0=0$.

ancient mathematician
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