Question

If Ln=limx→∞((x+14)(x+142)(x+143)……(x+14n))1n−x then the number of solution of equation (x−1x)(limn→∞(nLn))=|logex| is

Answer 1

$x=\frac{1}{t}$
$L_n=\underset{t\to 0}{\text{lim}}[{\left(\frac{(1+\frac{t}{4})(1+\frac{t}{4^2})\dots \dots (1+\frac{t}{4^n})}{t^n}\right)}^{\frac{1}{n}}-\frac{1}{t}]$
$y\left(t\right)={((1+\frac{t}{4})(1+\frac{t}{4^2})\dots \dots (1+\frac{t}{4^n}))}^{\frac{1}{n}}$
$lny\left(t\right)=\frac{1}{n}\{ln(1+\frac{t}{4})+ln(1+\frac{t}{4^2})+\dots \dots +ln(1+\frac{t}{4^n}\}$
$\frac{y^{{}^{\prime }}\left(t\right)}{y\left(t\right)}=\frac{1}{n}[\frac{1}{t+4}+\frac{1}{t+4^2}+\dots \dots +\left(\frac{1}{t+4^n}\right)]$
$L_n=\underset{t\to 0}{\text{lim}}\frac{1}{n}(\frac{1}{t+4}+\dots \dots +\frac{1}{t+4^n}){((1+\frac{t}{4})\dots \dots (1+\frac{t}{4^n}))}^{\frac{1}{n}}$
$L_n=\frac{\frac{1}{4}+\frac{1}{4^2}+\dots \dots +\frac{1}{4^n}}{n}=\frac{1}{n}\frac{\left(\frac{1}{4}\right)(1-{\left(\frac{1}{4}\right)}^n)}{(1-\frac{1}{4})}$
$\underset{n\to \infty }{\text{lim}}\frac{1-{\left(\frac{1}{4}\right)}^n}{3n}.n=\frac{1}{3}$
So $(x-\frac{1}{x})\left(\frac{1}{3}\right)=|{\text{log}}_{e}x|$ has two solutions