Calculating $\lim_{(n,x) \rightarrow (\infty, 1^+)}x^n$?

Question

Obviously, when $|x| \leq 1$, $\lim_{n \rightarrow \infty}x^n$ has a value such that $|x^n| \leq 1$, and $|x^n| \rightarrow \infty$ for $|x|>1$.

However, this led me to wonder, what would happen in the case that $x$ did not equal 1, but rather approached 1?

In other words, what would $$\lim_{(n,x) \rightarrow (\infty, 1^+)}x^n$$ be?

I am not that well-versed in mathematics (yet!), so it would be much appreciated if the answer could be explained simply as possible!

Answer 1

For any real number $a>1$, the sequence $x_n=\sqrt[n]{a}$ has the property that $x_n\to 1^+$, and $$\lim_{n\to\infty}(x_n)^n=\lim_{n\to\infty}a=a$$ So the limit you're asking about doesn't exist.

Answer 2

It depends on how $x$ approaches zero. Let $(x_n)$ be a sequence converging to $1$, write $y_n := n(x_n - 1)$, then $$ x_n^n = \left(1 + \frac{y_n}n\right)^n $$ which converges iff $y_n$ converges to some $y \in \mathbf R$, namely to $e^y$. So $x_n^n \to e^{\lim_n n(x_n - 1)}$ if $\lim n(x_n - 1)$ exists.

So, for example $$\left(1 + \frac 1{\sqrt n}\right)^n \to \infty, \quad \left(1 + \frac 1{n}\right)^n \to e, \quad \left(1 + \frac 1{n^2}\right)^n \to 1 $$ As we have different limits for different sequences, the limit $$ \lim_{(n,x) \to (\infty, 1^+)} x^n $$ does not exist.