$\infty$ is it?

We know that period of $\text{sin}^2x$ is $\pi$ and that of $f(kx)$ where k is a constant is $\large\displaystyle \frac{T}{k}$ where $\text{T}$ is the time period of $f(x)$ .

So, period of $\text{sin}^{2m}x$ is $\pi$ and that of $\large \text{sin}^{2m}x\sqrt{k}$ is $\large \dfrac{\pi}{\sqrt{k}}$

Since it is given that period is $\pi$ ,

$\large \dfrac{\pi}{\sqrt{k}}=\pi \Rightarrow k=1$

$\displaystyle \lim_{n\to∞}1^n=1$ .

Note that, it is not e because it is defined as $\displaystyle \large \lim_{x\to∞}(1+\frac{1}{x})^x=e$