Tuesday, 20 March 2018

limits - 1 to the power of infinity, why is it indeterminate?





I've been taught that $1^\infty$ is undetermined case. Why is it so? Isn't $1*1*1...=1$ whatever times you would multiply it? So if you take a limit, say $\lim_{n\to\infty} 1^n$, doesn't it converge to 1? So why would the limit not exist?


Answer



It isn’t: $\lim_{n\to\infty}1^n=1$, exactly as you suggest. However, if $f$ and $g$ are functions such that $\lim_{n\to\infty}f(n)=1$ and $\lim_{n\to\infty}g(n)=\infty$, it is not necessarily true that



$$\lim_{n\to\infty}f(n)^{g(n)}=1\;.\tag{1}$$



For example, $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e\approx2.718281828459045\;.$$



More generally,




$$\lim_{n\to\infty}\left(1+\frac1n\right)^{an}=e^a\;,$$



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



$$\lim_{n\to\infty}\left(1+\frac1n\right)^{n^2}=\infty\;,$$



and



$$\lim_{n\to\infty}\left(1+\frac1n\right)^{\sqrt n}=0\;,$$




so a limit of the form $(1)$ always has to be evaluated on its own merits; the limits of $f$ and $g$ don’t by themselves determine its value.


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...