Question

Rezolvarea va rog!!!!

Answer 1

Răspuns:

Ai raspunsul in poze

Explicație pas cu pas:

Answer 2

$l_1 = \lim\limits_{x\to 0}(\cos x\cdot \cos 2x\cdot ...\cdot \cos nx)^{\dfrac{1}{n^3x^2}}\\ \\ \ln(l_1) = \lim\limits_{x\to 0}\Big[\ln(\cos x\cdot \cos 2x\cdot ...\cdot \cos nx)^{\dfrac{1}{n^3x^2}}\Big] = \\ \\ = \lim\limits_{x\to 0}\ln\Big[\prod\limits_{k=1}^n {\cos(kx)\Big]}^{\dfrac{1}{n^3x^2}} =\lim\limits_{x\to 0}\dfrac{\ln\Big[\prod\limits_{k=1}^n {\cos(kx)\Big]}}{n^3x^2}=$

$=\lim\limits_{x\to 0} \dfrac{\sum\limits_{k=1}^n \ln(\cos kx)}{n^3x^2}=\lim\limits_{x\to 0} \dfrac{\sum\limits_{k=1}^n \dfrac{-k\sin(kx)}{\cos(kx)}}{2n^3x} = \\\\ = \lim\limits_{x\to 0} \dfrac{\sum\limits_{k=1}^n -k\tan(kx)}{2n^3x} =\lim\limits_{x\to 0} \dfrac{\sum\limits_{k=1}^n \dfrac{-k^2}{\cos^2 (kx)}}{2n^3}= \\ \\ =\dfrac{-\sum\limits_{k=1}^n k^2}{2n^3} =\dfrac{-\dfrac{n(n+1)(2n+1)}{6}}{2n^3} =\\ \\\\ = -\dfrac{2n^3+3n^2+n}{12n^3}$

$\Rightarrow l_1 = e^{-\dfrac{2n^3+3n^2+n}{12n^3}}\\ \\ \Rightarrow L =\lim\limits_{n\to \infty} e^{-\dfrac{2n^3+3n^2+n}{12n^3}} = e^{-\dfrac{2}{12}} = \boxed{e^{-\dfrac{1}{6}}}$