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While studying indeterminate forms I like many others had used this post to understand why $1^\infty$ is indeterminate. Today I thought of a new and perhaps simpler argument to explain why this is so but I couldn't find it among the answers in the post. My argument is that we can write $1^x = (-1)^{2.x}$
Now if $x = \infty$, then $1^\infty = (-1)^{\infty}$,
Which is only true if we can say infinity is even or odd which we can't say.
So, am I correct, or I have made some mistakes in my assumptions.

amWhy
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Madly_Maths
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    In the case of $1^{\infty}$ , the parity plays no role. We have $1^x=1$ for every real $x$. But $1^{\infty}$ is in fact indeterminate since for example $(1+x)^{1/x}$ tends to $e$ for $x\to 0$. $(-1)^x$ is not even defined for non-integer values. If we only take integer values , THEN your argument applies that $\infty$ is no number , neither even nor odd. – Peter May 19 '24 at 12:56
  • Also, the exponential laws only hold for positive real bases. – Peter May 19 '24 at 12:59
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    @Peter $-1^x = e^{i\pi x}$ is defined for non-integer values. – paulina May 19 '24 at 13:16
  • Without context , everyone would assume that we are in the real numbers. Also, exponential laws do not hold in general for non-real bases. – Peter May 19 '24 at 13:18
  • In general, $1^x \ne (-1)^{2x}$. – jjagmath May 19 '24 at 13:30
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    Although your argument doesn't bear up, I think you definitely deserve points for creativity, applied speculation, and curiosity. All valuable and admirable attributes for a mathematician. – fleablood May 19 '24 at 16:35
  • Thank you for the kind words – Madly_Maths May 19 '24 at 16:46
  • @jjagmath can you please tell why that is so in general? – Madly_Maths May 19 '24 at 16:47
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    @Madly_Maths In short, $((-1)^2)^x$ may not equal $(-1)^{2x}$. The usual rule of $(a^b)^c = a^{bc}$ fails pretty quickly after we go outside elementary context of $a>0$ or integer $c$. – Brian Moehring May 19 '24 at 18:42
  • @Madly_Maths Take $x=1/2$. Then $1^x=1^{1/2}=1$ while $(-1)^{2x}=(-1)^1 = -1$. – jjagmath May 19 '24 at 19:15
  • Thank you for the explanations. I understand. – Madly_Maths May 20 '24 at 04:05

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No, your argument is not correct. The issue of $1^\infty$ being an indeterminate form has nothing to do with integer exponents per se; one is instead interested in real valued exponents approaching $\infty$. It also has nothing to do with changing the base to $-1$, which on the infinitesmal scale of limits is far, far away from $1$; one is instead interested in real valued bases approaching $1$.

I suggest simply reading the accepted answer of the post you linked to. In brief summary, what matters are the values of the expression $b^x$ as one lets the base $b$ approach $1$ and the exponent $x$ approach $\infty$: the limiting value is not well defined. That's what makes the expression $1^\infty$ an indeterminate form.

Lee Mosher
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