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?
 A: There are many reasons. For example, let $1^\infty=1$. Taking logarithm, you have $\infty\cdot 0=0$. Similarly for other operations you will obtain some absurd.
A: 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.
A: The short answer is that it is because $x^y$ can tend to any nonnegative limit as $x\to1$ and $y\to\infty$. For one example, consider the classical limit $$\lim_{n\to\infty}\Bigl(1+\frac xn\Bigr)^n=e^x.$$
A: The limit of $1^{\infty}$ exist:$$\lim_{n\to\infty}1^n$$ is not indeterminate. However$$\lim_{a\to 1^+,n\to\infty}a^n$$ is indeterminate.. 
