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Is $x=\infty$ considered or not as a solution to $\exp(-x)=0$ ?

If not, why?

Aaron Hendrickson
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Mathieu Krisztian
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    It depends on the set where you are looking for solutions. If the set is $\mathbb{R}$, then the answer is no. – FormerMath Jun 10 '21 at 18:28
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    $\infty$ is not a number, therefore it cannot be a solution of the equation—which simply has no real solution. In fact, the notation $x=\infty$ is a little misleading; you should think in terms of limits: $\exp(-x) \to 0$ as $x\to\infty$. – ИΛJΛ Jun 10 '21 at 18:32
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    @Naja "$\infty$ is not a number" doesn't make sense. Which numbers are you claiming it is not one of? – plop Jun 10 '21 at 18:34

1 Answers1

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It is not a solution in the real numbers. ($\infty$ is not a real number.)

It is not a solution in the complex numbers. ($\infty$ is not a complex number.)

It is not a solution in the Riemann sphere $\overline{\mathbb C}$: Yes, $\infty \in \overline{\mathbb C}$, but $\exp(-x)$ has an essential singularity at $x=\infty$.

So it is best just to say $$ \lim_{x\to +\infty} \exp(-x) = 0 $$ and not $\exp(-\infty) = 0$.

An important property of the exponential function is $\exp(z) \ne 0$ for all $z$.

GEdgar
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