I"ve to be taught that $1^infty$ is undetermined case. Why is that so? Isn"t $1*1*1...=1$ whatever times you would multiply it? so if you take it a limit, to speak $lim_n oinfty 1^n$, doesn"t it converge to 1? for this reason why would the limit no exist?




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It isn’t: $lim_n oinfty1^n=1$, specifically as you suggest. However, if $f$ and also $g$ are functions such that $lim_n oinftyf(n)=1$ and $lim_n oinftyg(n)=infty$, the is not necessarily 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 together $a$ varieties over all genuine numbers, $e^a$ varieties over all confident 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 form $(1)$ always has to be evaluate on its very own merits; the limits of $f$ and also $g$ nothing by themselves identify its value.


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answer Mar 3 "13 at 20:29
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Brian M. ScottBrian M. Scott
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The limit of $1^infty$ exist:$$lim_n oinfty1^n$$ is no indeterminate. However$$lim_a o 1^+,n oinftya^n$$ is indeterminate..


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edited Apr 15 "18 at 4:43
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KingLogic
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There are numerous reasons. Because that example, let $1^infty=1$. Acquisition logarithm, you have $inftycdot 0=0$. An in similar way for various other operations you will attain some absurd.


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answered Mar 3 "13 at 20:25
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Boris NovikovBoris Novikov
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The brief answer is that it is since $x^y$ have the right to tend to any type of nonnegative limit as $x o1$ and also $y oinfty$. For one example, take into consideration the classical limit $$lim_n oinftyBigl(1+frac xnBigr)^n=e^x.$$


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edited Apr 13 "18 at 0:27
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Paul Sinclair
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answer Mar 3 "13 in ~ 20:20
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