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