You are watching: 1 to the power of infinity

It isn’t: $lim_n oinfty1^n=1$, specifically as friend suggest. However, if $f$ and $g$ are attributes such that $lim_n oinftyf(n)=1$ and $lim_n oinftyg(n)=infty$, the is

**not**have to true that

$$lim_n oinftyf(n)^g(n)=1;. ag1$$

For example, $$lim_n oinftyleft(1+frac1n ight)^n=eapprox2.718281828459045;.$$

More generally,

$$lim_n oinftyleft(1+frac1n ight)^an=e^a;,$$

and as $a$ ranges over all real numbers, $e^a$ varieties over all hopeful real numbers. Finally,

$$lim_n oinftyleft(1+frac1n ight)^n^2=infty;,$$

and

$$lim_n oinftyleft(1+frac1n ight)^sqrt n=0;,$$

so a border of the type $(1)$ constantly has to be evaluate on its very own merits; the boundaries of $f$ and also $g$ don’t by themselves determine its value.

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answered Mar 3 "13 in ~ 20:29

Brian M. ScottBrian M. Scott

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The limit of $1^infty$ exist:$$lim_n oinfty1^n$$ is not indeterminate. However$$lim_a o 1^+,n oinftya^n$$ is indeterminate..

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edited Apr 15 "18 in ~ 4:43

KingLogic

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answered Mar 3 "13 at 20:21

user35039user35039

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There are plenty of reasons. Because that example, permit $1^infty=1$. Taking logarithm, you have $inftycdot 0=0$. Likewise for various other operations you will acquire some absurd.

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answer Mar 3 "13 in ~ 20:25

Boris NovikovBoris Novikov

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The quick answer is that it is due to the fact that $x^y$ deserve to tend to any nonnegative limit together $x o1$ and $y oinfty$. For one example, consider the timeless limit $$lim_n oinftyBigl(1+frac xnBigr)^n=e^x.$$

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edited Apr 13 "18 at 0:27

Paul Sinclair

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answer Mar 3 "13 in ~ 20:20

Harald Hanche-OlsenHarald Hanche-Olsen

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