$1^∞$ in different aspects I was solving different limits when I noticed the following:
I solved the following limit(which is something less than 1 raised to power infinity) and found that it approaches to 1/e
$$\begin{align}\lim_{x\to 0} (1 - x)^{1/x}\end{align}$$
But I have also seen written that :
$$\lim_{n\to\infty} (\sin x)^n = 0 $$  when $0 < \sin x <1$ , but my question is will not $(\sin x)^\infty$ sometimes approach a value similar to upper limit?
 A: As the comments explained already, the reason you get a value different from $0$, $1$ or $+\infty$ in the first term is because the basis changes along with the exponent. In the second term $\sin x$ is a constant in regards to $n$ and for constants we have
$$\lim_{n\to\infty} c^n = \left\{\begin{array}{rl}+\infty&\mbox{if } c > 1,\\ 1&\mbox{if }c = 1,\\0&\mbox{if } 0\leq c < 1.\end{array}\right.$$
$(c^n)_{n\in\mathbb N}$ does not converge when $c<0$.
The story is different if we allow $c$ to be dependent on $n$. In fact, since for any $z\in\mathbb R$ we have
$$\lim_{n\to\infty} \left(1+\frac{z}{n}\right)^n=e^z$$
(note that the $z$ is a constant in regards to $n$, but $1+\frac{z}{n}$ is dependent on n) we can have any positive number $y$ as our result by choosing $z=\ln y$.
This is why the term $1^\infty$ doesn't make sense if you take the $1$ to mean a sequence that converges to $1$ and the infinity as a sequence that converges to $\infty$; it has uncountable possible values. It would be as undefined as $\frac{\infty}{\infty}$ or $\infty-\infty$ because similarily you could get any number if you mean these infinities to represent sequences converging to $\infty$.
