Most craziest limit of cosines

Considering the first couple of terms in the relevant Taylor series, we have $\big(\cos jx\big)^{\frac{1}{j}} \; \approx \; 1 - \tfrac12jx^2 + \cdots \hspace{2cm} \big(\cos jx\big)^j \; \approx \; 1 - \tfrac12j^3x^2 + \cdots$ and hence $\begin{aligned} \prod_{j=1}^n \big(\cos jx\big)^{\frac{1}{j}} & \approx 1 - \tfrac12\sum_{j=1}^n j \,x^2 + \cdots \; = \; 1 - \tfrac14n(n+1)x^2 + \cdots \\ \prod_{j=1}^n \big(\cos jx)\big)^j & \approx 1 - \frac12\sum_{j=1}^n j^3 \,x^2 \; = \; 1 - \tfrac18n^2(n+1)^2 x^2 + \cdots \end{aligned}$ and hence $\prod_{j=1}^n \big(\cos jx\big)^{\frac{1}{j}} - \prod_{j=1}^n \big(\cos jx)\big)^j \; \approx \; \big[\tfrac18n^2(n+1)^2 - \tfrac14n(n+1)\big]x^2 + \cdots \; = \; \tfrac18n(n+1)(n-1)(n+2)x^2 + \cdots$ while $\big(\cos nx)^{\frac{1}{n}} - \big(\cos nx\big)^n \; \approx \; \tfrac12(n^3 - n)x^2 + \cdots \; = \; \tfrac12n(n-1)(n+1)x^2 + \cdots$ and hence $\frac{\prod_{j=1}^n \big(\cos jx\big)^{\frac{1}{j}} - \prod_{j=1}^n \big(\cos jx)\big)^j }{\big(\cos nx)^{\frac{1}{n}} - \big(\cos nx\big)^n} \; \approx \; \tfrac14(n+2) + \cdots$ and hence this ratio tends to $\tfrac12(n+2)$ as $x \to 0+$ . With $n = 2016$ , the answer is $\boxed{504.5}$ .

We use Approximation Mclaurin Series we know that mclaurin series for $\displaystyle \cos \left( x\right)=\sum_{n=0}^{\infty} (-1)^{n}\frac{x^{2n}}{(2n)!}=1-\frac{x^{2}}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$

and too this problem we only need the first of 2 term. So $\displaystyle \cos \left( x\right) \approx 1-\frac{x^{2}}{2!}=1-\frac{x^{2}}{2}$

then limit become , $\large \displaystyle \lim_{x\rightarrow 0^{+}} \frac{\left( 1-\frac{x^2}{2} \right)\left( 1-\frac{4}{2}x^{2} \right)^{\frac{1}{2}} \left( 1-\frac{9}{2}x^{2} \right)^{\frac{1}{3}}\cdots \left( 1-\frac{2016^{2}}{2}x^{2} \right)^{\frac{1}{2016}} - \left( 1-\frac{x^2}{2} \right)\left( 1-\frac{4}{2}x^{2} \right)^{2} \left( 1-\frac{9}{2}x^{2} \right)^{3}\cdots \left( 1-\frac{2016}{2}x^{2} \right)^{2016} }{\left( 1-\frac{2016^{2}}{2}x^{2} \right)^{\frac{1}{2016}} -\left( 1-\frac{2016}{2}x^{2} \right)^{2016} }$ After that, we use Approximation Binomial Newton from that we know $\boxed{\displaystyle \left( 1+x \right)^{n}=\sum_{k=0}^{n} \binom{n}{k}x^{k} }$

then the approximation is we take first 2 term $\displaystyle \left( 1+x \right)^{n}\approx \binom{n}{0}+\binom{n}{1}x^{1}=1+nx$

Now See a denominator become : $\left( 1-\frac{2016}{2}x^{2} \right)^{\frac{1}{2016}} -\left( 1-\frac{2016}{2}x^{2} \right)^{2016} = \left( 1-\frac{2016}{2}x^{2} \right) -\left( 1-\frac{2016^{3}}{2}x^{2} \right) = \frac{2016^{3}-2016}{2}x^{2} = \frac{1}{2}\left( 2016 \right)\left( 2016^{2}-1 \right)x^{2}$

so the limit become $\displaystyle \lim_{x\rightarrow 0^{+}} \frac{\left( 1-\frac{1}{2}x^{2} \right)\left( 1-\frac{2}{2}x^{2} \right) \left( 1-\frac{3}{2}x^{2} \right)\cdots \left( 1-\frac{2016}{2}x^{2} \right) - \left( 1-\frac{1^3}{2}x^{2} \right)\left( 1-\frac{2^{3}}{2}x^{2} \right) \left( 1-\frac{3^3}{2}x^{2} \right)\cdots \left( 1-\frac{2016^{3}}{2}x^{2} \right) }{\frac{1}{2}\left( 2016 \right)\left( 2016^{2}-1 \right)x^{2} } \\ \displaystyle = \lim_{x\rightarrow 0^{+}} \frac{ \left( 1- \frac{1}{2}\left( 1+2+3+\cdots+2016 \right)x^{2}+ \left(1\times 2+1\times 3+ \cdots +2015\times 2016 \right)x^{4} +\cdots \right) - \left( 1- \frac{1}{2}\left( 1^{3}+2^{3}+3^{3}+\cdots+2016^{3} \right)x^{2}+ \left( 1^{3}2^{3}+1^{3}3^{3}+ \cdots+2015^{3}2016^{3} \right)x^{4} +\cdots \right) }{\frac{1}{2}\left( 2016 \right)\left( 2016^{2}-1 \right)x^{2} } \\ \displaystyle = \lim_{x\rightarrow 0^{+}} \frac{ \left( 1- \frac{1}{2}\left( \frac{1}{2}(2015)(2016) \right)x^{2}+ \left(1\times 2+1\times 3+ \cdots +2015\times 2016 \right)x^{4} +\cdots+\left( \cdots \right)x^{4032} \right) - \left( 1- \frac{1}{2}\left( \left( \frac{1}{2}\left( 2015 \right)\left( 2016 \right) \right)^{2} \right)x^{2}+ \frac{1}{4}\left( 1^{3}2^{3}+1^{3}3^{3}+ \cdots+2015^{3}2016^{3} \right)x^{4} +\cdots+\left( \cdots \right)x^{4032} \right) }{\frac{1}{2}\left( 2016 \right)\left( 2015 \right)\left( 2017 \right)x^{2} } \\ \displaystyle = \lim_{x\rightarrow 0^{+}} \frac{ \left( \frac{1}{2}\left( \left( \frac{1}{2}\left( 2015 \right)\left( 2016 \right) \right)^{2}-\frac{1}{2}(2015)(2016) \right)x^{2}+ \frac{1}{4}\left(1\times 2+1\times 3+ \cdots +2015\times 2016 - 1^{3}2^{3}+1^{3}3^{3}+ \cdots+2015^{3}2016^{3} \right)x^{4} +\cdots +\left( \cdots \right)x^{4032} \right) }{\frac{1}{2}\left( 2016 \right)\left( 2015 \right)\left( 2017 \right)x^{2} } \\ \displaystyle \lim_{x\rightarrow 0^{+}} \frac{ \left( \frac{1}{2}\left( \left( \frac{1}{2}\left( 2015 \right)\left( 2016 \right) \right)^{2}-\frac{1}{2}(2015)(2016) \right)x^{2}+ \frac{1}{4}\left(1\times 2+1\times 3+ \cdots +2015\times 2016 - 1^{3}2^{3}+1^{3}3^{3}+ \cdots+2015^{3}2016^{3} \right)x^{4} +\cdots +\left( \cdots \right)x^{4032} \right) }{\frac{1}{2}\left( 2016 \right)\left( 2015 \right)\left( 2017 \right)x^{2} }$

divide by $x^{2}$ for denominator and numerator. We have $\displaystyle \lim_{x\rightarrow 0^{+}} \frac{ \left( \frac{1}{2}\left( \left( \frac{1}{2}\left( 2016 \right)\left( 2017 \right) \right)^{2}-\frac{1}{2}(2016)(2017) \right)+ \frac{1}{4}\left(1\times 2+1\times 3+ \cdots +2015\times 2016 - 1^{3}2^{3}+1^{3}3^{3}+ \cdots+2015^{3}2016^{3} \right)x^{2} +\cdots +\left( \cdots \right)x^{4030} \right) }{\frac{1}{2}\left( 2016 \right)\left( 2015 \right)\left( 2017 \right) }$ subtitute for limit $x\rightarrow 0^{+}$ then the answer is

$\displaystyle \frac{ \left( \frac{1}{2}\left( \left( \frac{1}{2}\left( 2016 \right)\left( 2017 \right) \right)^{2}-\frac{1}{2}(2016)(2017) \right) \right) }{\frac{1}{2}\left( 2016 \right)\left( 2015 \right)\left( 2017 \right) } \\ = \displaystyle \frac{ \left( \frac{1}{2}\left( 2016 \right)\left( 2017 \right) \right)\left( \frac{1}{2}\left( 2016 \right)\left( 2017 \right)-1 \right)}{\left( 2016 \right)\left( 2015 \right)\left( 2017 \right) } \\ =\displaystyle \frac{1}{4}\frac{2016\times 2017-2}{2015}\\ =\displaystyle \frac{1}{4}\frac{(2015+1)(2015+2)-2}{2015}\\ =\displaystyle \frac{1}{4}\frac{2015^{2}+3\times 2015}{2015}\\ =\displaystyle \frac{2015+3}{4}\\ =\displaystyle \frac{2018}{4}=\boxed{504.5}$