lim_(x→0) ((1/x) − ((ln^(1000) (1 + x))/x^(1001) ))


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Question Number 34755 by Joel579 last updated on 10/May/18

$\lim_{x \to 0} (\frac{1}{x} - \frac{\ln^{1000} (1 + x)}{x^{1001}})$
Commented by Joel579 last updated on 10/May/18

$L = \lim_{x \to 0} \frac{1}{x} (1 - \frac{\ln^{1000} (1 + x)}{x^{1000}})$ $= \lim_{x \to 0} [\frac{1 - {(\frac{\ln (1 + x)}{x})}^{1000}}{x}]$ $= \lim_{x \to 0} - 1000 {(\frac{\ln (1 + x)}{x})}^{999} . [\frac{\frac{x}{1 + x} - \ln (1 + x)}{x^{2}}] L^{'} H o p i t a l$ $= \lim_{x \to 0} [- 1000 {(1 - \frac{x}{2} + \frac{x}{3} - . . .)}^{999}] . [\frac{x - x \ln (1 + x) - \ln (1 + x)}{x^{2} (1 + x)}]$ $= - 1000 . \lim_{x \to 0} [\frac{1}{1 + x}] [\frac{x - \ln (1 + x)}{x^{2}} - \frac{\ln (1 + x)}{x}]$ $= - 1000 . 1 . \lim_{x \to 0} [\frac{x - \ln (1 + x)}{x^{2}} - (1 - \frac{x}{2} + \frac{x}{3} - . . .)]$ $= - 1000 . 1 . [\frac{1}{2} - 1]$ $L = 500$ $\ln (1 + x) = - \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k} . x^{k}}{k}$ $= x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - . . .$ $\frac{\ln (1 + x)}{x} = \frac{1}{x} (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - . . .)$ $= 1 - \frac{x}{2} + \frac{x}{3} - . . .$
Commented by Joel579 last updated on 10/May/18

$M a y b e s o m e o n e h a s a n o t h e r m e t h o d ?$
Commented by abdo mathsup 649 cc last updated on 11/May/18

$l e t f i n d {l i m}_{x \to o} (\frac{1}{x} - \frac{{l n}^{a} (1 + x)}{x^{a + 1}}) w i t h a f r o m N$ $= {l i m}_{x \to 0} \frac{x^{a + 1} - x {l n}^{a} (1 + x)}{x^{a + 2}} b u t$ ${l n (1 + x))}^{^{'}} = \frac{1}{1 + x} = 1 - x + o (x^{2}) \Rightarrow$ $l n (1 + x) = x - \frac{x^{2}}{2} + o (x^{3}) \Rightarrow=$ ${(l n (1 + x))}^{a} = x^{a} {(1 - \frac{x}{2} + o (x))}^{a} \sim x^{a} (1 - \frac{a x}{2})$ $\Rightarrow \frac{x^{a + 1} - x {l n}^{a} (1 + x)}{x^{a + 2}} \sim \frac{x^{a + 1} - x^{a + 1} + \frac{{a x}^{a + 2}}{2}}{x^{a + 2}}$ $\sim \frac{a}{2} \Rightarrow {l i m}_{x \to 0} (\frac{1}{x} - \frac{{l n}^{a} (1 + x)}{x^{a + 1}}) = \frac{a}{2}$ $l e t t a k e a = 1000 \Rightarrow$ ${l i m}_{x \to 0} (\frac{1}{x} - \frac{{l n}^{1000} (1 + x)}{x^{1001}}) = 500 .$
Commented by Joel579 last updated on 11/May/18

$T h a n k y o u, b u t I {d i d n}^{'} t u n d e r s t a n d$ $w h a t o (x^{2}) i s$
Commented by abdo mathsup 649 cc last updated on 11/May/18

$o (x^{2}) = x^{2} ξ (x) w i t h {l i m}_{x \to 0} ξ (x) = 0$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/May/18
	$=((lim)/(x→0)) (1/x)[1−{((ln(1+x))/x)}^(1000) ] =((lim)/(x→0))(1/x)[1−{((x−(x^2 /2)+(x^3 /3)−(x^4 /4)+...)/x)}^(1000) ] =((lim)/(x→0))(1/(x ))[1−{1−(x/2)+(x^2 /3)−(x^3 /4)+...}^(1000) ] =((lim)/(x→0))(1/x)[1−{1−ψ(x)}] here ψ(x) are the terms containing x^(1000) and higher power of x... =((lim)/(x→0))(1/x)[ψ(x)] now {1−ψ(x)}={1−(x/2)+(x^2 /3)−(x^3 /4)+...}^(1000) ={1−(x/2)}^(1000) ignoring other terms ={1−1000_C_1 .(x/2)+other terms containg x^k to the power x^(2 ) and ablve...so ψ(x)=500x+other terms clntaining x^k k≥2 =((lim)/(x→0))((500x+other terms clintaing x^k )/x) =500$
	$= \frac{l i m}{x \to 0} \frac{1}{x} [1 - {\frac{l n (1 + x)}{x}}^{1000}]$ $= \frac{l i m}{x \to 0} \frac{1}{x} [1 - {\frac{x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + . . .}{x}}^{1000}]$ $= \frac{l i m}{x \to 0} \frac{1}{x} [1 - {1 - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + . . .}^{1000}]$ $= \frac{l i m}{x \to 0} \frac{1}{x} [1 - {1 - ψ (x)}]$ $h e r e ψ (x) a r e t h e t e r m s c o n t a i n i n g x^{1000} a n d$ $h i g h e r p o w e r o f x . . .$ $= \frac{l i m}{x \to 0} \frac{1}{x} [ψ (x)]$ $n o w {1 - ψ (x)} = {1 - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + . . .}^{1000}$ $= {1 - \frac{x}{2}}^{1000} i g n o r i n g o t h e r t e r m s$ $= {1 - 1000_{C_{1}} . \frac{x}{2} + o t h e r t e r m s c o n t a i n g x^{k} t o t h e$ $p o w e r x^{2} a n d a b l v e . . . s o ψ (x) = 500 x + o t h e r$ $t e r m s c l n t a i n i n g x^{k} k ⩾ 2$ $= \frac{l i m}{x \to 0} \frac{500 x + o t h e r t e r m s c l i n t a i n g x^{k}}{x}$ $= 500$
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