For $|x|\gt1,\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac2{x^{2^k}+x^{-2^k}}\right)=f(x)$ 
For $|x|\gt1,$ $$\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac2{x^{2^k}+x^{-2^k}}\right)=f(x)$$
Show that

*

*A) $\int_2^5f(x)dx=3+\ln16$

*B) $\lim_{x\to\infty}f(x)=1$

*C) $f(x)=0$ has one solution

*D) $f(x)$ is decreasing function in the domain of function.


$$f(x)=\lim_{n \to \infty}\prod_{k=0}^{n} \left(\frac{x^{2^k}+x^{-2^k}+2}{x^{2^k}+x^{-2^k}}\right)\\=\lim_{n \to \infty}\prod_{k=0}^{n}\frac{(x^{2^{k-1}}+x^{-2^{k-1}})^2}{x^{2^k}+x^{-2^k}}\\=\lim_{n\to\infty}\frac{(x^{2^{-1}}+x^{-2^{-1}})^2}{x^{2^0}+x^{-2^0}}\frac{(x^{2^{0}}+x^{-2^{0}})^2}{x^{2^1}+x^{-2^1}}\frac{(x^{2^{1}}+x^{-2^{1}})^2}{x^{2^2}+x^{-2^2}}...\frac{(x^{2^{n-1}}+x^{-2^{n-1}})^2}{x^{2^n}+x^{-2^n}}\\=\lim_{n\to\infty}(x^{2^{-1}}+x^{-2^{-1}})^2(x^{2^{0}}+x^{-2^{0}})(x^{2^{1}}+x^{-2^{1}})...\frac{(x^{2^{n-1}}+x^{-2^{n-1}})}{x^{2^n}+x^{-2^n}}$$
$x+\frac1x\gt2\implies x^2+\frac1{x^2}\gt2\implies x^4+\dfrac1{x^4}\gt2...\implies x^{2^n}+\dfrac1{x^{2^n}}\gt2$
Not able to proceed next.

Let $P$ denote the product, without the limit.
Note that:
$$\left(1+\frac {2}{x^{2^k}+\frac {1}{x^{2^k}}}\right)=\frac {\left(x^{2^{k-1}}+\frac {1}{x^{2^{k-1}}}\right)^2}{x^{2^k}+\frac {1}{x^{2^k}}}$$
Using this leads to a partial telescopic product. I shall skip the cancellation steps, and write directly the result as:
$$P=\frac {\left(\sqrt x+\frac {1}{\sqrt x}\right)^2}{x^{2^n}+\frac {1}{x^{2^n}}}\prod_{k=0}^{n-1} \left(x^{2^k}+\frac {1}{x^{2^k}}\right)$$
Let the remaining product be called $Q$, then multiplying and dividing $Q$ by $x-\frac 1x$, we get (again, simplification has been left for the reader):
$$Q=\frac {x^{2^n}-\frac {1}{x^{2^n}}}{x-\frac 1x}$$
Thus, we have:
$$f(x)=\frac {(\sqrt x+\frac {1}{\sqrt x})^2}{x-\frac 1x} \lim_{n\to \infty} \frac {x^{2^n}-\frac {1}{x^{2^n}}}{x^{2^n}+\frac {1}{x^{2^n}}}$$
The limit part evaluates to $1$, hence, after simplification we obtain:
$$f(x)=\frac {x+1}{x-1}$$
The rest is quite simple, all options shall be correct.
