lim_ _(x→∞) ((ln x)/x) = ? You can only use series expansion / sandwich theorem!


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Question Number 34843 by rahul 19 last updated on 11/May/18

$\underset{x \to \infty}{\lim_{}} \frac{\ln x}{x} = ?$ $Y o u c a n o n l y u s e$ $s e r i e s e x p a n s i o n / s a n d w i c h t h e o r e m!$
Commented by abdo mathsup 649 cc last updated on 11/May/18
$changement x =1 +(1/t) give ((lnx)/x) = ((ln(1 + (1/t)))/(1+(1/t))) = ((ln(1+t) −ln(t))/((t+1)/t)) =(t/(t+1)){ln(1+t) −ln(t)} = ((t ln(1+t) −tln(t))/(t+1)) = (t/(t+1))ln(1+t) −((tln(t))/(t+1)) but ln( 1+t)^. = (1/(1+t)) =Σ_(n=0) ^∞ (−1)^n t^n ln(1+t) = Σ_(n=0) ^∞ (((−1)^n )/(n+1))t^(n+1) = Σ_(n=1) ^∞ (((−1)^(n−1) )/n) t^n ((ln(x))/x) = (t/(t+1)) ( t −(t^2 /2) +(t^3 /3) +...) −t ln(t)Σ_(n=0) ^∞ (−1)^n t^n = (t^2 /(t+1))( 1−(t/2) +(t^2 /3)+...) −Σ_(n=0) ^∞ (−1)^n t^(n+1) ln(t)→0 when t→0 so lim_(x→+∞) ((lnx)/x) =0$
$c h a n g e m e n t x = 1 + \frac{1}{t} g i v e$ $\frac{l n x}{x} = \frac{l n (1 + \frac{1}{t})}{1 + \frac{1}{t}} = \frac{l n (1 + t) - l n (t)}{\frac{t + 1}{t}}$ $= \frac{t}{t + 1} {l n (1 + t) - l n (t)} = \frac{t l n (1 + t) - t l n (t)}{t + 1}$ $= \frac{t}{t + 1} l n (1 + t) - \frac{t l n (t)}{t + 1} b u t$ $l n {(1 + t)}^{.} = \frac{1}{1 + t} = \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n}$ $l n (1 + t) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} t^{n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} t^{n}$ $\frac{l n (x)}{x} = \frac{t}{t + 1} (t - \frac{t^{2}}{2} + \frac{t^{3}}{3} + . . .) - t l n (t) \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n}$ $= \frac{t^{2}}{t + 1} (1 - \frac{t}{2} + \frac{t^{2}}{3} + . . .) - \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n + 1} l n (t) \to 0$ $w h e n t \to 0 s o {l i m}_{x \to + \infty} \frac{l n x}{x} = 0$
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