Question Number 34755 by Joel579 last updated on 10/May/18
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{{x}}\:−\:\frac{\mathrm{ln}^{\mathrm{1000}} \:\left(\mathrm{1}\:+\:{x}\right)}{{x}^{\mathrm{1001}} }\right) \\ $$
Commented by Joel579 last updated on 10/May/18
![L = lim_(x→0) (1/x)(1 − ((ln^(1000) (1 + x))/x^(1000) )) = lim_(x→0) [((1 − (((ln (1 + x))/x))^(1000) )/x)] = lim_(x→0) −1000(((ln (1 + x))/x))^(999) . [(((x/(1 + x)) − ln (1 + x))/x^2 )] L′Hopital = lim_(x→0) [−1000(1 − (x/2) + (x/3) − ...)^(999) ] . [((x − x ln (1 + x) − ln (1 + x))/(x^2 (1 + x)))] = −1000 . lim_(x→0) [(1/(1 + x))][((x − ln (1 + x))/x^2 ) − ((ln (1 + x))/x)] = −1000 . 1 . lim_(x→0) [((x − ln (1 + x))/x^2 ) − (1 − (x/2) + (x/3) − ...)] = −1000 . 1 . [(1/2) − 1] L = 500 ln (1 + x) = −Σ_(k=1) ^∞ (((−1)^k . x^k )/k) = x − (x^2 /2) + (x^3 /3) − ... ((ln (1 + x))/x) = (1/x)(x − (x^2 /2) + (x^3 /3) − ...) = 1 − (x/2) + (x/3) − ...](https://www.tinkutara.com/question/Q34761.png)
$${L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}}\left(\mathrm{1}\:−\:\frac{\mathrm{ln}^{\mathrm{1000}} \:\left(\mathrm{1}\:+\:{x}\right)}{{x}^{\mathrm{1000}} }\right) \\ $$$$\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}\:−\:\left(\frac{\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}\right)}{{x}}\right)^{\mathrm{1000}} }{{x}}\right] \\ $$$$\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:−\mathrm{1000}\left(\frac{\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}\right)}{{x}}\right)^{\mathrm{999}} \:.\:\left[\frac{\frac{{x}}{\mathrm{1}\:+\:{x}}\:−\:\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}\right)}{{x}^{\mathrm{2}} }\right]\:\:\:{L}'{Hopital} \\ $$$$\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[−\mathrm{1000}\left(\mathrm{1}\:−\:\frac{{x}}{\mathrm{2}}\:+\:\frac{{x}}{\mathrm{3}}\:−\:…\right)^{\mathrm{999}} \right]\:.\:\left[\frac{{x}\:−\:{x}\:\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}\right)\:−\:\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}\:+\:{x}\right)}\right] \\ $$$$\:\:\:\:\:=\:−\mathrm{1000}\:.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{1}\:+\:{x}}\right]\left[\frac{{x}\:−\:\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}\right)}{{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}\right)}{{x}}\right] \\ $$$$\:\:\:\:\:=\:−\mathrm{1000}\:.\:\mathrm{1}\:.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{{x}\:−\:\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}\right)}{{x}^{\mathrm{2}} }\:−\:\left(\mathrm{1}\:−\:\frac{{x}}{\mathrm{2}}\:+\:\frac{{x}}{\mathrm{3}}\:−\:…\right)\right] \\ $$$$\:\:\:\:\:=\:−\mathrm{1000}\:.\:\mathrm{1}\:.\:\left[\frac{\mathrm{1}}{\mathrm{2}}\:−\:\mathrm{1}\right] \\ $$$${L}\:=\:\mathrm{500} \\ $$$$ \\ $$$$\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}\right)\:=\:−\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{k}} \:.\:{x}^{{k}} }{{k}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{x}\:−\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\:\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:−\:… \\ $$$$\frac{\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}\right)}{{x}}\:=\:\frac{\mathrm{1}}{{x}}\left({x}\:−\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\:\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:−\:…\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}\:−\:\frac{{x}}{\mathrm{2}}\:+\:\frac{{x}}{\mathrm{3}}\:−\:… \\ $$
Commented by Joel579 last updated on 10/May/18
$${Maybe}\:{someone}\:{has}\:{another}\:{method}\:? \\ $$
Commented by abdo mathsup 649 cc last updated on 11/May/18
$${let}\:{find}\:\:{lim}_{{x}\rightarrow{o}} \left(\frac{\mathrm{1}}{{x}}\:−\frac{{ln}^{{a}} \left(\mathrm{1}+{x}\right)}{{x}^{{a}+\mathrm{1}} }\:\right)\:\:{with}\:{a}\:{from}\:{N} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{x}^{{a}+\mathrm{1}} \:\:−{x}\:{ln}^{{a}} \left(\mathrm{1}+{x}\right)}{{x}^{{a}+\mathrm{2}} }\:\:{but} \\ $$$$\left.{ln}\left(\mathrm{1}+{x}\right)\right)^{'} \:=\:\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\mathrm{1}−{x}\:+{o}\left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}\right)\:=\:{x}\:\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right)\Rightarrow= \\ $$$$\left({ln}\left(\mathrm{1}+{x}\right)\right)^{{a}} \:\:=\:{x}^{{a}} \left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\:+{o}\left({x}\right)\right)^{{a}} \:\sim{x}^{{a}} \left(\mathrm{1}−\frac{{ax}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\:\frac{{x}^{{a}+\mathrm{1}} \:−{x}\:{ln}^{{a}} \left(\mathrm{1}+{x}\right)}{{x}^{{a}+\mathrm{2}} }\:\sim\:\frac{{x}^{{a}+\mathrm{1}} \:\:−{x}^{{a}+\mathrm{1}} \:+\frac{{ax}^{{a}+\mathrm{2}} }{\mathrm{2}}}{{x}^{{a}+\mathrm{2}} } \\ $$$$\sim\:\frac{{a}}{\mathrm{2}}\:\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}}{{x}}\:−\frac{{ln}^{{a}} \left(\mathrm{1}+{x}\right)}{{x}^{{a}\:+\mathrm{1}} }\:\right)\:=\:\frac{{a}}{\mathrm{2}} \\ $$$${let}\:{take}\:\:{a}\:=\mathrm{1000}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}}{{x}}\:−\frac{{ln}^{\mathrm{1000}} \left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{1001}} }\right)\:=\mathrm{500}. \\ $$
Commented by Joel579 last updated on 11/May/18
$${Thank}\:{you},\:{but}\:{I}\:{didn}'{t}\:{understand} \\ $$$${what}\:{o}\left({x}^{\mathrm{2}} \right)\:{is} \\ $$
Commented by abdo mathsup 649 cc last updated on 11/May/18
$${o}\left({x}^{\mathrm{2}} \right)={x}^{\mathrm{2}} \xi\left({x}\right)\:\:{with}\:{lim}_{{x}\rightarrow\mathrm{0}} \xi\left({x}\right)=\mathrm{0} \\ $$
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](https://www.tinkutara.com/question/Q34874.png)
$$=\frac{{lim}}{{x}\rightarrow\mathrm{0}}\:\frac{\mathrm{1}}{{x}}\left[\mathrm{1}−\left\{\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\right\}^{\mathrm{1000}} \right] \\ $$$$=\frac{{lim}}{{x}\rightarrow\mathrm{0}}\frac{\mathrm{1}}{{x}}\left[\mathrm{1}−\left\{\frac{{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+…}{{x}}\right\}^{\mathrm{1000}} \right] \\ $$$$=\frac{{lim}}{{x}\rightarrow\mathrm{0}}\frac{\mathrm{1}}{{x}\:}\left[\mathrm{1}−\left\{\mathrm{1}−\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{{x}^{\mathrm{3}} }{\mathrm{4}}+…\right\}^{\mathrm{1000}} \right] \\ $$$$=\frac{{lim}}{{x}\rightarrow\mathrm{0}}\frac{\mathrm{1}}{{x}}\left[\mathrm{1}−\left\{\mathrm{1}−\psi\left({x}\right)\right\}\right] \\ $$$${here}\:\psi\left({x}\right)\:{are}\:{the}\:{terms}\:{containing}\:{x}^{\mathrm{1000}} \:{and}\: \\ $$$${higher}\:{power}\:{of}\:{x}… \\ $$$$=\frac{{lim}}{{x}\rightarrow\mathrm{0}}\frac{\mathrm{1}}{{x}}\left[\psi\left({x}\right)\right] \\ $$$${now}\:\left\{\mathrm{1}−\psi\left({x}\right)\right\}=\left\{\mathrm{1}−\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{{x}^{\mathrm{3}} }{\mathrm{4}}+…\right\}^{\mathrm{1000}} \\ $$$$=\left\{\mathrm{1}−\frac{{x}}{\mathrm{2}}\right\}^{\mathrm{1000}} \:{ignoring}\:{other}\:{terms} \\ $$$$=\left\{\mathrm{1}−\mathrm{1000}_{{C}_{\mathrm{1}} } .\frac{{x}}{\mathrm{2}}+{other}\:{terms}\:{containg}\:{x}^{{k}} {to}\:{the}\right. \\ $$$${power}\:{x}^{\mathrm{2}\:} {and}\:{ablve}…{so}\:\psi\left({x}\right)=\mathrm{500}{x}+{other} \\ $$$${terms}\:{clntaining}\:{x}^{{k}} \:\:{k}\geqslant\mathrm{2}\: \\ $$$$=\frac{{lim}}{{x}\rightarrow\mathrm{0}}\frac{\mathrm{500}{x}+{other}\:{terms}\:{clintaing}\:{x}^{{k}} \:}{{x}} \\ $$$$ \\ $$$$=\mathrm{500}\: \\ $$