November 9th, 2017, 01:29 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 232 Thanks: 26  Limit with sequence
How to find this limit? $\displaystyle \lim _{x\rightarrow 0}x^{1}(1\prod \limits _{j=1}^{n}\sqrt[j]{\cos jx}) =$ $\displaystyle \lim _{x\rightarrow 0} \frac{1\cos x \cdot \sqrt{\cos 2x}\cdot \sqrt[3]{\cos 3x}\cdot ...\cdot \sqrt[n]{\cos nx}}{x}=\,?$ Last edited by skipjack; November 9th, 2017 at 02:34 AM. 
November 9th, 2017, 02:34 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,162 Thanks: 1638 
What is the limit for n = 1? Can you find the limit for n = 2?

November 9th, 2017, 04:31 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra 
I get a limit of zero using Taylor expansions. \begin{align*} \left. \begin{aligned} \cos{(x)} &= 1  x^2 + \mathcal{O}(x^2) &\implies&& \cos{(nx)} &= 1  (nx)^2 + \mathcal{O}(x^2) \\[8pt] &&&&\sqrt[n]{1 + x} &= 1 + \tfrac1nx + \mathcal{O}(x) \end{aligned} \right\} \implies \sqrt[n]{\cos{(nx)}} &= 1  \tfrac1n(nx)^2 + \mathcal{O}(x^2) \\ &= 1  nx^2 + \mathcal{O}(x^2) \end{align*} Thus the product \begin{align*} \prod_{n=1}^\infty \sqrt[n]{\cos{(nx)}} &= \prod_{n=1}^\infty \left(1  nx^2 + \mathcal{O}(x^2) \right) \\ &= \left(1  x^2 + \mathcal{O}(x^2) \right)\left(1  2x^2 + \mathcal{O}(x^2) \right)\left(1  3x^2 + \mathcal{O}(x^2) \right)\ldots \\ &= 1 + \mathcal{O}(x) & \text{(for some constant $a$)} \end{align*} by expanding the product term by term. Notice that, at this point we are assuming that the product has a finite value$^*$. We can motivate this by pointing out that for $x \lt \frac\pi2$, every term in the product is between $0$ and $1$, and thus the product must also be between $0$ and $1$. Now we can put this into the limit expression \begin{align*} \lim_{x \to 0} \frac{1  \prod_{n=1}^\infty \sqrt[n]{\cos{(nt)}}}{x} &= \lim_{x \to 0} \frac{1  \big(1 + \mathcal{O}(x)\big)}{x} \\ &= \lim_{x \to 0} \frac{ \mathcal{O}(x)}{x} \\ &= 0 \end{align*} https://www.desmos.com/calculator/b4rtdppucm $^*$ Note that the coefficient of $x^2$ is actually $\left(\sum n\right)$ over the natural numbers. I assume that this leads to use of $\sum n = \frac1{12}$. Last edited by v8archie; November 9th, 2017 at 05:19 AM. 
November 9th, 2017, 09:17 AM  #4 
Senior Member Joined: Dec 2015 From: Earth Posts: 232 Thanks: 26 
$\displaystyle \lim_{x \to 0} \frac{1  \prod_{n=1}^\infty \sqrt[n]{\cos{(nt)}}}{x}=\lim_{x \to 0} \frac{ e^{\ln \prod_{n=1}^\infty \sqrt[n]{\cos{(nt)}}}1}{x}=\lim_{x \to 0}\sum _{n=1}^{\infty}(nx)^{1}\ln cos(nt)=0$

November 9th, 2017, 09:50 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra 
I'm not convinced at all by that second equals sign.


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