If period of $\sin^{2m} \sqrt{k}x$, $m\in \mathbb{N}$ is $\pi$ then find $\lim \limits_{n\to \infty }k^{n}$.
My attempt:
Period of $$\sin^{2m} \sqrt{k}x=\frac{\pi}{\sqrt{k}}$$
According to question
$$\frac{\pi}{\sqrt{k}}=\pi$$
$$k=1$$
Therefore $$\lim \limits_{n\to \infty }k^{n}= \lim \limits_{n\to \infty }1^{n}$$
Which is $1^{\infty}$ form.
I think it's answer should be ''Indeterminate" but Wolfram Alpha is saying that answer is $1$.
Can anyone please explain what's the correct answer?
"Indeterminate" is not a number, and it is never an answer to "what is this limit?" The limit either exists (and then it is a real number, or possibly $\pm\infty$), or it doesn't.
What "indeterminate" means here is neither more nor less than if you have a limit of the form $$ \lim_{n\to\infty} f(n)^{g(n)} $$ and it turns out that $\lim_{n\to\infty} f(n) = 1$ and $\lim_{n\to\infty} g(n)=\infty$, then knowing the limits of $f$ and $g$ is not enough to tell you what the value of the original limit is. It does not mean that the original limit doesn't exist (which it may or may not), and it certainly doesn't mean that the original limit is "indeterminate". All it means is that you need to find it by some diffferent method than taking limits of $f$ and $g$ separately.
In the case of $\lim_{n\to\infty} 1^n$ it is true that the expression asks for a limit of indeterminate form, meaning that computing $\lim_{n\to\infty}1$ and $\lim_{n\to\infty} n$ are not going to get you anywhere. So you need something else, but the "something else" easy enough: Each term in the sequence equals $1$, and therefore $1$ is also the limit.