Undefined reference $1^{\infty}$ in a limit such as $\lim_{n\to\infty}(1+1^n)$

Question

If $1^{\infty}$ is undefined reference in a limit, how is $$\lim\limits_{n\to\infty}(1+1^n)=2$$

Answer 1

$$\lim\limits_{n\to\infty}(1+1^n)=\lim\limits_{n\to\infty}(1+1)=2$$ as $$1^n=1$$ for any $n$

Answer 2

$0^0$, $\infty^0$, $1^\infty$ are "undefined in general", that is, they can assume different values case by case.

If you have a specific case (like this), you can establish a value for them, by applying the rules of limit.

Answer 3

It is more common (I think) to say: $1^\infty$ is an indeterminate form. Knowing only $\lim f(n) = 1$ and $\lim g(n) = \infty$, we cannot determine $\lim\;(f(n)^{g(n)})$.

In your case, $\lim\;(1^n) = 1$. But in other cases we can have things like $\lim\;(1+\frac{1}{n})^n = e$.

Answer 4

Clearly,$$\lim\limits_{n\to\infty}(1+1^n)=1+\lim\limits_{n\to\infty}1^n,$$let's ignore the first term.

Then $\lim_{n\to\infty}1^n$ is not an indeterminate form $1^\infty$, it is just $1$, as $1^n=1$ for all $n$.

The sequence is $$1,1,1,1,1,1,1,1,1,1,1\cdots$$and there is no doubt.

An undeterminate form $1^\infty$ requires "the $1$ not to be exactly $1$", like $\lim_{n\to\infty}\left(1+\dfrac1n\right)^n$, such that $1+\dfrac1n\ne1$.

The sequence is $$2,\frac94,\frac{64}{27},\frac{625}{256},\frac{7776}{3125},\frac{117649}{46656}\cdots$$ and it takes more effort to evaluate the limit, $e=2.718281828459\cdots$.