$$\lim_{n\to \infty} \{(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n})\}= f(x) \text{ for } |x|<1, \text{ then } f(\frac 12)= \text{?}$$
I need a hint to solve this problem because I am unable to proceed in any way with it.
I took logarithm on both sides of $f(x)$ equation and then tried but that didn't help too. I can see that gradually $x\to 0$
Hint:
Multiply and divide the expression by $1-x$
And use $a^2-b^2=(a-b)(a+b)$
EDIT: Multiplication and division gives
$$\lim_{n\to \infty} \frac{(1-x)(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n})}{1-x}= f(x)$$
$$=\lim_{n \to \infty}\frac{(1-x^2)(1+x^2)(1+x^4)....(1+x^{2^n})}{1-x}$$Again use $$a^2-b^2=(a+b)(a-b)$$ which gives
$$=\lim_{n \to \infty}\frac{(1-x^4)(1+x^4)....(1+x^{2^n})}{1-x}$$ Continuing like this gives
$$f(x)=\lim_{n \to \infty}\frac{1-x^{2^{n+1}}}{1-x}$$
Now what can you conclude if the number $|x|<1$ is raised to an infinitely big power..