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

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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
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
answer Mar 3 "13 in ~ 20:20
Harald Hanche-OlsenHarald Hanche-Olsen
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