Menu Close

Question-80585

Tinku Tara 0 Comments
Pin




Question Number 80585 by ahmadshahhimat775@gmail.com last updated on 04/Feb/20
Commented by kaivan.ahmadi last updated on 04/Feb/20
2. lim_(x→(π/6)) ((2sin2x+cosx)/(2sin2x−3cosx))=(((√3)+((√3)/2))/( (√3)−((3(√3))/2)))=(((3(√3))/2)/((−(√3))/2))=−3
$$\mathrm{2}. \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{6}}} \:\frac{\mathrm{2}{sin}\mathrm{2}{x}+{cosx}}{\mathrm{2}{sin}\mathrm{2}{x}−\mathrm{3}{cosx}}=\frac{\sqrt{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\:\sqrt{\mathrm{3}}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}{\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}}}=−\mathrm{3} \\ $$
Commented by kaivan.ahmadi last updated on 04/Feb/20
5. lim_(x→(π/4)) (tgx−1)tg2x=lim_(x→(π/4)) ((tgx−1)/(cot2x))= lim_(x→(π/4)) ((1+tg^2 x)/(−2(1+cot^2 2x)))=((1+1)/(−2(1+0)))=−1
$$\mathrm{5}. \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:\left({tgx}−\mathrm{1}\right){tg}\mathrm{2}{x}={lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:\frac{{tgx}−\mathrm{1}}{{cot}\mathrm{2}{x}}= \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}+{tg}^{\mathrm{2}} {x}}{−\mathrm{2}\left(\mathrm{1}+{cot}^{\mathrm{2}} \mathrm{2}{x}\right)}=\frac{\mathrm{1}+\mathrm{1}}{−\mathrm{2}\left(\mathrm{1}+\mathrm{0}\right)}=−\mathrm{1} \\ $$
Commented by kaivan.ahmadi last updated on 04/Feb/20
6. lim_(x→e) ((lnx+1−1)/(2/x))=((1+1−1)/(2/e))=(e/2)
$$\mathrm{6}. \\ $$$${lim}_{{x}\rightarrow{e}} \frac{{lnx}+\mathrm{1}−\mathrm{1}}{\frac{\mathrm{2}}{{x}}}=\frac{\mathrm{1}+\mathrm{1}−\mathrm{1}}{\frac{\mathrm{2}}{{e}}}=\frac{{e}}{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 04/Feb/20
1) let A_n =((2^n (n!))/n^n ) we have n!∼ n^n e^(−n) (√(2πn)) ⇒ A_n ∼((2^n .n^n e^(−n) (√(2π))n)/n^n ) =((2/e))^n (√(2πn)) =(√(2π))e^(nln((2/e))) ×e^((1/2)ln(n)) =(√(2π))×e^(nln((2/e))+ln((√n))) but we hsve nln((2/e))+ln((√n)) =nln(2)−n +(1/2)ln(n) =n{ln(2)−1+((ln(n))/(2n))}→+∞ ⇒ lim_(n→+∞) A_n =+∞
$$\left.\mathrm{1}\right)\:{let}\:{A}_{{n}} =\frac{\mathrm{2}^{{n}} \left({n}!\right)}{{n}^{{n}} }\:\:{we}\:{have}\:{n}!\sim\:{n}^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\:\Rightarrow \\ $$$${A}_{{n}} \sim\frac{\mathrm{2}^{{n}} .{n}^{{n}} {e}^{−{n}} \sqrt{\mathrm{2}\pi}{n}}{{n}^{{n}} }\:=\left(\frac{\mathrm{2}}{{e}}\right)^{{n}} \sqrt{\mathrm{2}\pi{n}}\:=\sqrt{\mathrm{2}\pi}{e}^{{nln}\left(\frac{\mathrm{2}}{{e}}\right)} ×{e}^{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}\right)} \\ $$$$=\sqrt{\mathrm{2}\pi}×{e}^{{nln}\left(\frac{\mathrm{2}}{{e}}\right)+{ln}\left(\sqrt{{n}}\right)} \:\:{but}\:{we}\:{hsve} \\ $$$${nln}\left(\frac{\mathrm{2}}{{e}}\right)+{ln}\left(\sqrt{{n}}\right)\:={nln}\left(\mathrm{2}\right)−{n}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}\right) \\ $$$$={n}\left\{{ln}\left(\mathrm{2}\right)−\mathrm{1}+\frac{{ln}\left({n}\right)}{\mathrm{2}{n}}\right\}\rightarrow+\infty\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =+\infty \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 04/Feb/20
forgive ln(2)−1<0 ⇒n{ln(2)−1+((ln(n))/(2n))}→−∞ ⇒ lim_(n→∞) A_n =0
$${forgive}\:{ln}\left(\mathrm{2}\right)−\mathrm{1}<\mathrm{0}\:\Rightarrow{n}\left\{{ln}\left(\mathrm{2}\right)−\mathrm{1}+\frac{{ln}\left({n}\right)}{\mathrm{2}{n}}\right\}\rightarrow−\infty\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} \:{A}_{{n}} =\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 04/Feb/20
2)changement sinx=t give lim_(x→(π/6)) ((2sin^2 x+sinx−1)/(2sin^2 x−3sinx+1)) =lim_(t→(1/2)) ((2t^2 +t−1)/(2t^2 −3t +1)) 2t^2 +t−1 =2(t−(1/2))(t−a)=(2t−1)(t−a)⇒a=−1 ⇒ 2t^2 +t−1 =(2t−1)(t+1) 2t^2 −3t +1 =2(t−(1/2))(t−b) =(2t−1)(t−b) ⇒b=1 ⇒ 2t^2 −3t +1 =(2t−1)(t−1) ⇒ lim_(t→(1/2)) (...) =lim_(t→(1/2)) (((2t−1)(t+1))/((2t−1)(t−1))) =lim_(t→(1/2)) ((t+1)/(t−1)) =((3/2)/(−(1/2)))=−3
$$\left.\mathrm{2}\right){changement}\:{sinx}={t}\:{give} \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{6}}} \:\:\:\frac{\mathrm{2}{sin}^{\mathrm{2}} {x}+{sinx}−\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{2}} {x}−\mathrm{3}{sinx}+\mathrm{1}}\:={lim}_{{t}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{\mathrm{2}{t}^{\mathrm{2}} \:+{t}−\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}\:+\mathrm{1}} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} +{t}−\mathrm{1}\:=\mathrm{2}\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({t}−{a}\right)=\left(\mathrm{2}{t}−\mathrm{1}\right)\left({t}−{a}\right)\Rightarrow{a}=−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2}{t}^{\mathrm{2}} +{t}−\mathrm{1}\:=\left(\mathrm{2}{t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right) \\ $$$$\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}\:+\mathrm{1}\:=\mathrm{2}\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({t}−{b}\right)\:=\left(\mathrm{2}{t}−\mathrm{1}\right)\left({t}−{b}\right)\:\Rightarrow{b}=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2}{t}^{\mathrm{2}} −\mathrm{3}{t}\:+\mathrm{1}\:=\left(\mathrm{2}{t}−\mathrm{1}\right)\left({t}−\mathrm{1}\right)\:\Rightarrow \\ $$$${lim}_{{t}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}} \:\left(…\right)\:={lim}_{{t}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\frac{\left(\mathrm{2}{t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)}{\left(\mathrm{2}{t}−\mathrm{1}\right)\left({t}−\mathrm{1}\right)}\:={lim}_{{t}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}\:=\frac{\frac{\mathrm{3}}{\mathrm{2}}}{−\frac{\mathrm{1}}{\mathrm{2}}}=−\mathrm{3} \\ $$
Commented by mathmax by abdo last updated on 04/Feb/20
let X_n =((n!)/4^n ) we have n!∼n^(n ) e^(−n) (√(2πn)) ⇒ X_n ∼ (n^n /((4e)^n ))(√(2π))n =e^(nln(n)) e^(−nln(4e)) (√(2π))e^((1/2)ln(n)) =e^(nln(n)−nln(4e)+((ln(n))/2)) =e^(n{ln(n)−ln(4e)+((ln(n))/(2n))}) →+∞
$${let}\:{X}_{{n}} =\frac{{n}!}{\mathrm{4}^{{n}} }\:\:{we}\:{have}\:{n}!\sim{n}^{{n}\:} {e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\:\Rightarrow \\ $$$${X}_{{n}} \sim\:\frac{{n}^{{n}} }{\left(\mathrm{4}{e}\right)^{{n}} }\sqrt{\mathrm{2}\pi}{n}\:\:={e}^{{nln}\left({n}\right)} {e}^{−{nln}\left(\mathrm{4}{e}\right)} \sqrt{\mathrm{2}\pi}{e}^{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}\right)} \\ $$$$={e}^{{nln}\left({n}\right)−{nln}\left(\mathrm{4}{e}\right)+\frac{{ln}\left({n}\right)}{\mathrm{2}}} \:={e}^{{n}\left\{{ln}\left({n}\right)−{ln}\left(\mathrm{4}{e}\right)+\frac{{ln}\left({n}\right)}{\mathrm{2}{n}}\right\}} \:\rightarrow+\infty \\ $$
Commented by mathmax by abdo last updated on 04/Feb/20
5) let f(x)=(tanx)^(tan(2x)) ⇒f(x)=e^(tan(2x)ln(tanx)) changement x=(π/4)+t give x→(π/4) ⇔t→0 f(x)=g(t)=e^(tan(2((π/4)+t))ln(tan((π/4)+t))) =e^(tan(2t+(π/2))ln(tan((π/4)+t))) we have tan(2t+(π/2))=((sin(2t+(π/2)))/(cos(2t+(π/2)))) =−(1/(tan(2t))) ⇒ tan((π/4)+t) =((1+tan(t))/(1−tan(t))) ⇒ln(tan((π/4)+t))=ln(1+tant)−ln(1−tant) ⇒tan(2t+(π/2))ln(tan((π/4)+t))∼((−1)/(2t))×(t−(−t)) =−1 ⇒ lim_(t→0) g(t)=(1/e) =lim_(x→(π/4))
$$\left.\mathrm{5}\right)\:{let}\:{f}\left({x}\right)=\left({tanx}\right)^{{tan}\left(\mathrm{2}{x}\right)} \:\Rightarrow{f}\left({x}\right)={e}^{{tan}\left(\mathrm{2}{x}\right){ln}\left({tanx}\right)} \\ $$$${changement}\:{x}=\frac{\pi}{\mathrm{4}}+{t}\:{give}\:\:{x}\rightarrow\frac{\pi}{\mathrm{4}}\:\Leftrightarrow{t}\rightarrow\mathrm{0} \\ $$$${f}\left({x}\right)={g}\left({t}\right)={e}^{{tan}\left(\mathrm{2}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right){ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right)} \\ $$$$={e}^{{tan}\left(\mathrm{2}{t}+\frac{\pi}{\mathrm{2}}\right){ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right)} \:\:{we}\:{have} \\ $$$${tan}\left(\mathrm{2}{t}+\frac{\pi}{\mathrm{2}}\right)=\frac{{sin}\left(\mathrm{2}{t}+\frac{\pi}{\mathrm{2}}\right)}{{cos}\left(\mathrm{2}{t}+\frac{\pi}{\mathrm{2}}\right)}\:=−\frac{\mathrm{1}}{{tan}\left(\mathrm{2}{t}\right)}\:\Rightarrow \\ $$$${tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\:=\frac{\mathrm{1}+{tan}\left({t}\right)}{\mathrm{1}−{tan}\left({t}\right)}\:\Rightarrow{ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right)={ln}\left(\mathrm{1}+{tant}\right)−{ln}\left(\mathrm{1}−{tant}\right) \\ $$$$\Rightarrow{tan}\left(\mathrm{2}{t}+\frac{\pi}{\mathrm{2}}\right){ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right)\sim\frac{−\mathrm{1}}{\mathrm{2}{t}}×\left({t}−\left(−{t}\right)\right)\:=−\mathrm{1}\:\Rightarrow \\ $$$${lim}_{{t}\rightarrow\mathrm{0}} \:\:\:{g}\left({t}\right)=\frac{\mathrm{1}}{{e}}\:={lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \\ $$
Commented by mathmax by abdo last updated on 04/Feb/20
lim_(x→(π/4)) f(x)=(1/e)
$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:\:{f}\left({x}\right)=\frac{\mathrm{1}}{{e}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *