find-lim-x-0-2-1-a-x-1-x-

Question Number 29689 by mrW2 last updated on 11/Feb/18

$${find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{2}}{\mathrm{1}+{a}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} =? \\ $$

Commented by abdo imad last updated on 13/Feb/18

let put A(x)=((2/(1+a^x )))^(1/x) ⇒ln(A(x))= (1/x)ln( (2/(1+a^x ))) =(1/x)( ln(2)−ln(1+a^x ))=(1/x)(ln(2)−ln(2 +a^x −1)) =(1/x)(−ln(1+((a^x −1)/2)))⇒A(x)= e^(((−1)/x)ln(1 +((a^x −1)/2))) for x∈v(0) ((a^x −1)/2) ∈v(0) and ln(1+u) ∼u ⇒ln(1+((a^x −1)/2))∼((a^x −1)/2)⇒ −(1/x)ln(...) ∼ ((1−a^x )/(2x)) by hospital theorem lim_(x→0) ((1−a^x )/(2x))=lim_(x→0) ((1−e^(xlna) )/(2x))=lim_(x→0) ((−lna e^(xlna) )/2) =ln((1/( (√a)))) ⇒lim_(x→0) A(x)= (1/( (√a))) .

$${let}\:{put}\:{A}\left({x}\right)=\left(\frac{\mathrm{2}}{\mathrm{1}+{a}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} \:\:\Rightarrow{ln}\left({A}\left({x}\right)\right)=\:\frac{\mathrm{1}}{{x}}{ln}\left(\:\frac{\mathrm{2}}{\mathrm{1}+{a}^{{x}} }\right) \\ $$$$=\frac{\mathrm{1}}{{x}}\left(\:{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{1}+{a}^{{x}} \right)\right)=\frac{\mathrm{1}}{{x}}\left({ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{2}\:+{a}^{{x}} −\mathrm{1}\right)\right) \\ $$$$=\frac{\mathrm{1}}{{x}}\left(−{ln}\left(\mathrm{1}+\frac{{a}^{{x}} −\mathrm{1}}{\mathrm{2}}\right)\right)\Rightarrow{A}\left({x}\right)=\:{e}^{\frac{−\mathrm{1}}{{x}}{ln}\left(\mathrm{1}\:+\frac{{a}^{{x}} −\mathrm{1}}{\mathrm{2}}\right)} \:\:{for}\:{x}\in{v}\left(\mathrm{0}\right) \\ $$$$\frac{{a}^{{x}} −\mathrm{1}}{\mathrm{2}}\:\in{v}\left(\mathrm{0}\right)\:{and}\:{ln}\left(\mathrm{1}+{u}\right)\:\sim{u}\:\Rightarrow{ln}\left(\mathrm{1}+\frac{{a}^{{x}} −\mathrm{1}}{\mathrm{2}}\right)\sim\frac{{a}^{{x}} −\mathrm{1}}{\mathrm{2}}\Rightarrow \\ $$$$−\frac{\mathrm{1}}{{x}}{ln}\left(…\right)\:\sim\:\frac{\mathrm{1}−{a}^{{x}} }{\mathrm{2}{x}}\:{by}\:{hospital}\:{theorem} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}−{a}^{{x}} }{\mathrm{2}{x}}={lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−{e}^{{xlna}} }{\mathrm{2}{x}}={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{−{lna}\:{e}^{{xlna}} }{\mathrm{2}} \\ $$$$={ln}\left(\frac{\mathrm{1}}{\:\sqrt{{a}}}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{A}\left({x}\right)=\:\frac{\mathrm{1}}{\:\sqrt{{a}}}\:. \\ $$

Commented by mrW2 last updated on 13/Feb/18

$${thanks}\:{sir}! \\ $$

Answered by ajfour last updated on 11/Feb/18

let lim_(x→0) ((2/(1+a^x )))^(1/x) = L ln L =lim_(x→0) [((ln (1+((1−a^x )/(1+a^x ))))/x)] =lim_(x→0) [(((((1−a^x )/(1+a^x ))))/x)] =lim_(x→0) (((−xln a−(x^2 /2)(ln a)^2 −...))/(x(1+a^x ))) =lim_(x→0) [−ln a(((1+((xln a)/2)+..)/(1+a^x )))] ⇒ ln L = −((ln a)/2) ⇒ L = (1/( (√a))) .

$${let}\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{1}+{a}^{{x}} }\right)^{\mathrm{1}/{x}} \:=\:{L} \\ $$$$\mathrm{ln}\:{L}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}−{a}^{{x}} }{\mathrm{1}+{a}^{{x}} }\right)}{{x}}\right] \\ $$$$\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\left(\frac{\mathrm{1}−{a}^{{x}} }{\mathrm{1}+{a}^{{x}} }\right)}{{x}}\right]\: \\ $$$$\:\:\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(−{x}\mathrm{ln}\:{a}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{ln}\:{a}\right)^{\mathrm{2}} −…\right)}{{x}\left(\mathrm{1}+{a}^{{x}} \right)} \\ $$$$\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[−\mathrm{ln}\:{a}\left(\frac{\mathrm{1}+\frac{{x}\mathrm{ln}\:{a}}{\mathrm{2}}+..}{\mathrm{1}+{a}^{{x}} }\right)\right] \\ $$$$\Rightarrow\:\:\mathrm{ln}\:{L}\:=\:−\frac{\mathrm{ln}\:{a}}{\mathrm{2}}\: \\ $$$$\Rightarrow\:\:\:\:\:\:\boldsymbol{{L}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{a}}}}\:\:. \\ $$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

are you guys teacher?.... you solve all ques easily..... or a dtudent of +1 +2?

$$\mathrm{are}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{teacher}?….\:\mathrm{you}\:\mathrm{solve}\:\mathrm{all}\:\mathrm{ques}\:\mathrm{easily}…..\:\mathrm{or}\:\mathrm{a}\:\mathrm{dtudent}\:\mathrm{of}\:+\mathrm{1}\:+\mathrm{2}? \\ $$

Commented by ajfour last updated on 11/Feb/18

i teach physics for 10+1,10+2 . what about you sir ?

$${i}\:{teach}\:{physics}\:{for}\:\mathrm{10}+\mathrm{1},\mathrm{10}+\mathrm{2}\:.\:{what}\:{about}\:{you}\:{sir}\:? \\ $$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

$$\mathrm{im}\:\mathrm{a}\:+\mathrm{1}\:\mathrm{student} \\ $$

Commented by math solver last updated on 11/Feb/18

wow! still solving maths q. efficiently. my phy. teacher should learn then!

$$\mathrm{wow}!\:\mathrm{still}\:\mathrm{solving}\:\mathrm{maths}\:\mathrm{q}.\:\mathrm{efficiently}. \\ $$$$\mathrm{my}\:\mathrm{phy}.\:\mathrm{teacher}\:\mathrm{should}\:\mathrm{learn}\:\mathrm{then}! \\ $$

Commented by math solver last updated on 11/Feb/18

by reading the answers of mrw2 sir i assumed he is a professor at some university but later i came to know he is an engineer and have very little (almost no) use of phy.,maths.

$$\mathrm{by}\:\mathrm{reading}\:\mathrm{the}\:\mathrm{answers}\:\mathrm{of}\:\mathrm{mrw2}\:\mathrm{sir} \\ $$$$\mathrm{i}\:\mathrm{assumed}\:\:\mathrm{he}\:\mathrm{is}\:\mathrm{a}\:\mathrm{professor}\:\mathrm{at}\:\mathrm{some}\: \\ $$$$\mathrm{university}\:\mathrm{but}\:\:\mathrm{later}\:\mathrm{i}\:\mathrm{came}\:\mathrm{to}\:\mathrm{know} \\ $$$$\mathrm{he}\:\mathrm{is}\:\mathrm{an}\:\mathrm{engineer}\:\mathrm{and}\:\mathrm{have}\:\mathrm{very}\:\mathrm{little} \\ $$$$\left(\mathrm{almost}\:\mathrm{no}\right)\:\mathrm{use}\:\mathrm{of}\:\mathrm{phy}.,\mathrm{maths}. \\ $$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

$$\mathrm{are}\:\mathrm{u}\:\mathrm{also}\:\mathrm{teacher}? \\ $$

Commented by math solver last updated on 11/Feb/18

no, i am also +1 student. moving to +2 this april.

$$\mathrm{no},\:\mathrm{i}\:\mathrm{am}\:\mathrm{also}\:+\mathrm{1}\:\mathrm{student}. \\ $$$$\mathrm{moving}\:\mathrm{to}\:+\mathrm{2}\:\mathrm{this}\:\mathrm{april}. \\ $$

Commented by math solver last updated on 11/Feb/18

$$\mathrm{just}\:\mathrm{go}\:\mathrm{to}\:\mathrm{question}\:\mathrm{ID}\:\mathrm{22537}. \\ $$

Commented by mrW2 last updated on 11/Feb/18

Thank you ajfour sir! I guess the answer should be 1/√a, see Q29633, but I had no proof. Now it is sure.

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

is ur target iit?... u f vry intelligent at 16 im sldo 16

$$\mathrm{is}\:\mathrm{ur}\:\mathrm{target}\:\mathrm{iit}?…\:\mathrm{u}\:\mathrm{f}\:\mathrm{vry}\:\mathrm{intelligent}\:\mathrm{at}\:\mathrm{16}\:\:\mathrm{im}\:\mathrm{sldo}\:\mathrm{16} \\ $$

Commented by math solver last updated on 11/Feb/18

yes ! btw, i am sorry i am a nerd infront of these guys asking silly questions :(

$$\mathrm{yes}\:! \\ $$$$\mathrm{btw},\:\mathrm{i}\:\mathrm{am}\:\mathrm{sorry}\:\mathrm{i}\:\mathrm{am}\:\mathrm{a}\:\mathrm{nerd}\:\mathrm{infront}\:\mathrm{of} \\ $$$$\mathrm{these}\:\mathrm{guys}\:\mathrm{asking}\:\mathrm{silly}\:\mathrm{questions}\::\left(\right. \\ $$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

$$\mathrm{thn}\:\mathrm{think}\:\mathrm{about}\:\mathrm{me}…\:\mathrm{i}\:\mathrm{also}\:\mathrm{target}\:\mathrm{iit} \\ $$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

fir bhi?...avg?... me abt 6 hrs generally

$$\mathrm{fir}\:\mathrm{bhi}?…\mathrm{avg}?…\:\mathrm{me}\:\mathrm{abt}\:\mathrm{6}\:\mathrm{hrs}\:\mathrm{generally} \\ $$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

$$\mathrm{im}\:\mathrm{in}\:+\mathrm{1moving}\:+\mathrm{2}\:\mathrm{this}\:\mathrm{april} \\ $$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

$$\mathrm{how}\:\mathrm{many}\:\mathrm{hours}\:\mathrm{a}\:\mathrm{day}\:\mathrm{u}\:\mathrm{study}? \\ $$

Commented by math solver last updated on 11/Feb/18

its not fixed , even sometimes it goes like i ′nt studied for a whole day .

$$\mathrm{its}\:\mathrm{not}\:\mathrm{fixed}\:,\:\mathrm{even}\:\mathrm{sometimes}\:\mathrm{it}\:\mathrm{goes} \\ $$$$\mathrm{like}\:\mathrm{i}\:'\mathrm{nt}\:\mathrm{studied}\:\mathrm{for}\:\mathrm{a}\:\mathrm{whole}\:\mathrm{day}\:. \\ $$

Commented by 803jaideep@gmail.com last updated on 11/Feb/18

im general category... im from punjab patiala by d way

$$\mathrm{im}\:\mathrm{general}\:\mathrm{category}…\:\mathrm{im}\:\mathrm{from}\: \\ $$$$\mathrm{punjab}\:\mathrm{patiala}\:\mathrm{by}\:\mathrm{d}\:\mathrm{way} \\ $$

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