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Let $f(s)$ and $g(s)$ be functions of complex $s$.

$\lim_{s \to 0} f(s)$ is defined and $\lim_{s \to 0} f(s) \ne 0$.

But $\lim_{s \to 0} g(s) = \infty$.

Does this still hold:

$$\lim_{s \to 0} f(s)g(s) = \lim_{s \to 0} f(s) \lim_{s \to 0} g(s)$$

To me it seems like this holds because both sides are $\infty$. But the 3rd comment at Is $ \lim_{n \to \infty} (n \cdot 0) = 0 $ or undefined? says that the product formula works only if both limits exist. That is like saying, if both limits do not exist, then the product formula does not work.

But I think in the above case, both limits do not exist and still the product formula works. What am I missing?

Math Learner
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  • If they are real valued functions, yes, the limit exist. It is $\pm \infty$ depending on the sign of $\lim_{s \to 0} f(s)$. – Kavi Rama Murthy Mar 03 '23 at 00:05
  • The difficulty is that there is a general convention that since $\infty$ is not a number, but is instead a symbol of unbounded growth, the expression $\lim_x \to a ~f(x) = \infty$ actually implies that the limit, as $x \to a$ of $f(x)$ does not exist. So, in the example that you cited, the $\displaystyle ~\lim_{s \to 0} \left[f(x) \times g(x)\right]~$ does not exist, and $\displaystyle ~\left[\lim_{s \to 0} f(x)\right] \times \left[ \lim_{s\to 0} g(x)\right]~$ does not exist. – user2661923 Mar 03 '23 at 00:10
  • Re my previous comment, I don't actually disagree with the comment of @geetha290krm. The (somewhat arbitrary) convention that I was taught was that if the limit exists, it must be finite. However, the convention of asserting that the limit can actually equal $\infty$ is also in common use. – user2661923 Mar 03 '23 at 00:14
  • See a discussion on extended limit laws: https://math.stackexchange.com/q/2971122/72031 – Paramanand Singh Mar 03 '23 at 03:20
  • The question has several statements about $\lim_{s \to 0} f(s).$ Was one of them supposed to be about $\lim_{s \to 0} g(s)$? – David K Mar 03 '23 at 04:20

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