finding limit of form $1^\infty$ $$f(n) := \lim_{x \to 0}\left(\big(1+\sin\frac{x}{2}\big)\big(1+\sin\frac{x}{2^2}\big) \cdots \big(1+\sin\frac{x}{2^n}\big)\right)^{1/x} $$
Find $\lim_{n \to \infty}f(n)$.
I used the formula $$\large\lim_{x\to a} f(x)^{g(x)} = e^{\large\lim_{x \to a} (f(x)-1)* g(x)}$$ where $$\large\lim_{x \to a} f(x) = 1 $$ and $$\large\lim_{x \to a} g(x) = \infty$$
so the expression was simplified to
$$f(n) = e^ {\frac{1}{x} \large\lim_{x \to 0}((1+\sin\frac{x}{2})(1+sin\frac{x}{2^2}) \cdots (1+sin\frac{x}{2^n})-1)}$$
$$f(n) = e^ {\frac{1}{x} \large\lim_{x \to 0}\biggl(1+ \bigl(\sin\frac{x}{2}+ \sin\frac{x}{2^2} \cdots +\sin\frac{x}{2^n}\bigl) + \bigl( \sin\frac{x}{2}.\sin\frac{x}{2^2} \cdots \bigl) + \cdots -1 \biggl)}$$
I am not able to further simplify it.
 A: You can use a different strategy, namely to compute
$$
a_n=\lim_{x\to0}\log\left(\left(\bigl(1+\sin\frac{x}{2}\bigr)\bigl(1+\sin\frac{x}{2^2}\bigr) \dots \bigl(1+\sin\frac{x}{2^n}\bigr)\right)^{1/x}\right)
$$
that can be rewritten as
$$
\lim_{x\to0}\sum_{k=1}^n\frac{\log(1+\sin(x/2^k))}{x}
$$
Now it's much easier, because
$$
\lim_{x\to0}\frac{\log(1+\sin(x/2^k))}{x}=\lim_{x\to0}\frac{x/2^k+o(x)}{x}=\frac{1}{2^k}
$$
Therefore
$$
a_n=\sum_{k=1}^n\frac{1}{2^k}=1-\frac{1}{2^{k}}
$$
Now your limit is
$$
\lim_{n\to\infty}e^{a_n}=e
$$
A: You are proceeding correctly. The important thing to note here is that $n$ can be considered a finite number, allowing us to evaluate $f(n)$. After that the limit on $n$ can be imposed. Hence, in your expression:
$$L=e^{\lim_{x\to 0} \frac {\sum_{r=1}^n \sin(\frac {x}{2^r}) +O(x^2)}{x}}$$
Here, since all terms, which are finite in number, other than the summation of sines lead to terms of degree $\geq 2$, they can be ignored. Thus, since $\lim_{x\to 0} \frac{\sin(\frac {x}{2^r})}{x}=2^{-r}$, we get:
$$L=e^{2^{-1}+2^{-2}+...+2^{-n}}=e^{1-(\frac 12)^n}$$
Thus, $f(n)=e^{1-(\frac 12)^n}$, and $\lim_{n\to \infty} f(n)=e$.
A: Fix $k\in\Bbb N$ and recall that $\sin t\sim t$ as $t\to0$. Then,
$$\lim_{x\to0}\left(1+\sin(x/2^k)\right)^{1/x}=\lim_{x\to0}\left[\left(1+\sin(x/2^k)\right)^{1/\sin(x/2^k)}\right]^{\sin(x/2^k)/x}= \rm Exp\left(\lim_{x\to0}\frac{\sin(x/2^k)}{x} \right) =\rm Exp\left(\lim_{x\to0}\frac{x/2^k}{x}\right)=e^{1/2^k}.
$$
Now,
$$f(n) := \lim_{x \to 0}\left(\big(1+\sin\frac{x}{2}\big)\big(1+\sin\frac{x}{2^2}\big) \cdots \big(1+\sin\frac{x}{2^n}\big)\right)^{1/x}=\prod_{k=1}^n \lim_{x\to0}\left(1+\sin(x/2^k)\right)^{1/x}=\prod_{k=1}^n e^{1/2^k}=\rm Exp\left(\sum_{k=1}^n\frac{1}{2^k}\right)
$$
So
$$ \lim_{n\to\infty}f(n)=\rm Exp\left(\sum_{k=1}^\infty\frac{1}{2^k}\right) = \rm Exp\left(\frac{1/2}{1-1/2}\right)=e.
$$
