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?

You are watching: What is 1 to the power of infinity

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.

re-superstructure
mention
follow
answer Mar 3 "13 at 20:29

Brian M. ScottBrian M. Scott
\$endgroup\$
include a comment |
10
\$egingroup\$
The limit of \$1^infty\$ exist:\$\$lim_n oinfty1^n\$\$ is no indeterminate. However\$\$lim_a o 1^+,n oinftya^n\$\$ is indeterminate..

re-superstructure
point out
follow
edited Apr 15 "18 at 4:43

KingLogic
reply Mar 3 "13 at 20:21
user35039user35039
\$endgroup\$
include a comment |
5
\$egingroup\$
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.

re-publishing
point out
monitor
answered Mar 3 "13 at 20:25

Boris NovikovBoris Novikov
\$endgroup\$
1
3
\$egingroup\$
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.\$\$

share
mention
follow
edited Apr 13 "18 at 0:27

Paul Sinclair
answer Mar 3 "13 in ~ 20:20
Harald Hanche-OlsenHarald Hanche-Olsen
\$endgroup\$
0

## Not the prize you're looking for? Browse various other questions tagged limits infinity indexes or asking your very own question.

Featured on Meta
166
Why is \$1^infty\$ considered to it is in an indeterminate form
4
Confusion over \$a_k = (1+frac1k)^k = e\$
connected
3
Why is \$fracsum_n=1^infty nsum_n=1^infty n\$ indeterminate?
1
limit on 2 variables approaching infinity
2
examining limits at optimistic and an unfavorable infinity
2
display the border of a function doesn't exist
4
Evaluate limit of the series: \$lim_n oinfty left\$
1
The Power legislation of boundaries at infinity
hot Network questions an ext hot questions
mmsanotherstage2019.com
company
ridge Exchange Network
site architecture / logo © 2021 stack Exchange Inc; user contributions license is granted under cc by-sa. Rev2021.10.25.40556

See more: There Are Years That Ask Questions And Years That Answer Meaning

mmsanotherstage2019.comematics ridge Exchange works finest with JavaScript enabled