Question Number 51005 by Tawa1 last updated on 23/Dec/18 | ||
Commented by maxmathsup by imad last updated on 23/Dec/18 | ||
$${let}\:{A}\left(\theta\right)\:=\frac{{sin}\left(\theta−\frac{\pi}{\mathrm{6}}\right)}{\sqrt{\mathrm{3}}\:−\mathrm{2}{cos}\theta}\:\:{changement}\:\:\theta\:−\frac{\pi}{\mathrm{6}}\:={x}\:{give} \\ $$$${A}\left(\theta\right)\:=\frac{{sinx}}{\sqrt{\mathrm{3}}−\mathrm{2}{cos}\left({x}+\frac{\pi}{\mathrm{6}}\right)}\:=\frac{{sinx}}{\sqrt{\mathrm{3}}−\mathrm{2}\left({cosx}\:{cos}\left(\frac{\pi}{\mathrm{6}}\right)−{sinx}\:{sin}\left(\frac{\pi}{\mathrm{6}}\right)\right)} \\ $$$$=\:\frac{{sinx}}{\sqrt{\mathrm{3}}−\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cos}−\frac{\mathrm{1}}{\mathrm{2}}{sinx}\right)}\:=\frac{{sinx}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}\:+{sinx}}\:{but}\:{lim}_{\theta\rightarrow\frac{\pi}{\mathrm{6}}} \:\:={lim}_{{x}\rightarrow\mathrm{0}} \frac{{sinx}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}+{sinx}} \\ $$$${sinx}\:\sim\:{x}\:\:\:{and}\:{cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:\frac{{sinx}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}\:−{sinx}}\:\sim\:\frac{{x}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)−{x}} \\ $$$$=\:\frac{{x}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}^{\mathrm{2}} −{x}}\:=\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}−\mathrm{1}}\:\rightarrow−\mathrm{1}\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{lim}_{\theta\rightarrow\frac{\pi}{\mathrm{6}}} \:\:\:{A}\left(\theta\right)\:=−\mathrm{1}\:\:\:\:\:\:. \\ $$$${error}\:{of}\:{typo}\:\:\frac{{sinx}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}\:+{sinx}}\:\sim\:\frac{{x}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)+{x}}\:=\frac{{x}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}^{\mathrm{2}} \:+{x}}\:= \\ $$$$=\:\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\:+\mathrm{1}}\:\rightarrow\mathrm{1}\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow\bigstar{lim}_{\theta\rightarrow\frac{\pi}{\mathrm{6}}} \:\:\:{A}\left(\theta\right)\:=\:\mathrm{1}\:\bigstar \\ $$ | ||
Commented by maxmathsup by imad last updated on 23/Dec/18 | ||
$${let}\:{A}\left({x}\right)\:=\frac{\mathrm{1}−{cosx}\sqrt{{cos}\left(\mathrm{2}{x}\right)}}{{x}^{\mathrm{2}} }\:\:\:{we}\:{have}\:{x}\:\in{V}\left(\mathrm{0}\right) \\ $$$${cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\:{and}\:\:{cos}\left(\mathrm{2}{x}\right)\sim\:\mathrm{1}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:\Rightarrow\sqrt{{cos}\left(\mathrm{2}{x}\right)}\sim\sqrt{\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\sim\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{2}{x}^{\mathrm{2}} \right)\:=\mathrm{1}−{x}^{\mathrm{2}} \:\Rightarrow{A}\left({x}\right)\sim\:\frac{\mathrm{1}−\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\right)}{{x}^{\mathrm{2}} }\:=\frac{{x}^{\mathrm{2}} \:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\rightarrow\frac{\mathrm{3}}{\mathrm{2}}\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:{A}\left({x}\right)\:=\frac{\mathrm{3}}{\mathrm{2}}\:. \\ $$ | ||
Commented by Tawa1 last updated on 24/Dec/18 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18 | ||
$$\left.\mathrm{2}\right){t}=\theta−\frac{\pi}{\mathrm{6}} \\ $$$${li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\:\frac{{sin}\left({t}\right)}{\sqrt{\mathrm{3}}\:−\mathrm{2}{cos}\left({t}+\frac{\pi}{\mathrm{6}}\right)} \\ $$$$={li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\:\frac{{sint}}{\sqrt{\mathrm{3}}\:−\mathrm{2}\left({cost}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{sint}.\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$={li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\:\frac{{sint}}{\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{3}}\:{cost}+{sint}} \\ $$$$={li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\:\frac{\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}{cos}\frac{{t}}{\mathrm{2}}}{\sqrt{\mathrm{3}}\:×\mathrm{2}{sin}^{\mathrm{2}} \frac{{t}}{\mathrm{2}}+\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}{cos}\frac{{t}}{\mathrm{2}}} \\ $$$$={li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\:\frac{{cos}\frac{{t}}{\mathrm{2}}}{\sqrt{\mathrm{3}}\:{sin}\frac{{t}}{\mathrm{2}}+{cos}\frac{{t}}{\mathrm{2}}}=\mathrm{1} \\ $$$${or}\:{method} \\ $$$${li}\underset{\theta\rightarrow\frac{\pi\:}{\mathrm{6}}} {{m}}\:\:\frac{{sin}\left(\theta−\frac{\pi}{\mathrm{6}}\right)}{\mathrm{2}\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{cos}\left(\theta\right)\right]}\:\:\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}={cos}\frac{\pi}{\mathrm{6}}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{li}\underset{\theta\rightarrow\frac{\pi\:}{\mathrm{6}}} {{m}}\:\frac{\mathrm{2}{sin}\left(\frac{\theta−\frac{\pi}{\mathrm{6}}}{\mathrm{2}}\right){cos}\left(\frac{\theta−\frac{\pi}{\mathrm{6}}}{\mathrm{2}}\right)}{\mathrm{2}{sin}\left(\frac{\frac{\pi}{\mathrm{6}}+\theta}{\mathrm{2}}\right){sin}\left(\frac{\theta−\frac{\pi}{\mathrm{6}}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{1} \\ $$ | ||
Commented by Tawa1 last updated on 23/Dec/18 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$ | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18 | ||
$${most}\:{welcome}... \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18 | ||
$$\left.\mathrm{3}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{1}−{cosx}\sqrt{{cos}\mathrm{2}{x}}}{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−{cos}^{\mathrm{2}} {x}\left(\frac{\mathrm{1}−{tan}^{\mathrm{2}} {x}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+{cosx}\sqrt{{cos}\mathrm{2}{x}}\:\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sec}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}}{{sec}^{\mathrm{2}} {x}×{x}^{\mathrm{2}} ×\left(\mathrm{1}+{cosx}\sqrt{{cos}\mathrm{2}{x}}\:\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}−{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sin}^{\mathrm{2}} {x}.\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)+{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{sinx}}{{x}}\right)^{\mathrm{2}} ×\left(\frac{\mathrm{2}+{cos}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$ | ||
Commented by Tawa1 last updated on 23/Dec/18 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18 | ||
$$\left.\mathrm{1}\right)\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left\{{sec}\left({x}+{y}\right)−{secx}\right\}+{ysec}\left({x}+{y}\right)}{{y}} \\ $$$$={x}\left[\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sec}\left({x}+{y}\right)−{secx}}{{y}}\right]+{secx} \\ $$$$={x}\left[\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{cosx}−{cos}\left({x}+{y}\right)}{{ycosxcos}\left({x}+{y}\right)}\right]+{secx} \\ $$$$={x}\left[\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{sin}\left({x}+\frac{{y}}{\mathrm{2}}\right){sin}\left(\frac{{y}}{\mathrm{2}}\right)}{{cosxcos}\left({x}+{y}\right)×\frac{{y}}{\mathrm{2}}×\mathrm{2}}\right]+{secx} \\ $$$$={x}×\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}+{secx} \\ $$$$={xtanxsecx}+{secx} \\ $$ | ||
Commented by Tawa1 last updated on 23/Dec/18 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$ | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18 | ||
$${most}\:{welcome}... \\ $$ | ||