Question Number 14336 by tawa tawa last updated on 30/May/17
Commented by ajfour last updated on 30/May/17
![cos x=1−(1−cos x) =1−2sin^2 ((x/2))=1−t lim_(x→0) (cos x)=lim_(x→0) [(1−t)^(−1/t) ]^(−tcot 2x) =lim_(x→0) (e^(−tcot 2x) )=lim_(x→0) e^(−y) lim_(x→0) y=lim_(x→0) [tcot 2x=(2sin^2 (x/2))(((cos 2x)/(2sin (x/3)cos (x/2)))) so lim_(x→0) y=0 hence lim_(x→0) (cos x)^(cot 2x) =e^(−lim_(x→0) y) =1 . (perhaps only right hand llimit ) .](https://www.tinkutara.com/question/Q14339.png)
$$\mathrm{cos}\:{x}=\mathrm{1}−\left(\mathrm{1}−\mathrm{cos}\:{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{1}−{t} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cos}\:{x}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\left(\mathrm{1}−{t}\right)^{−\mathrm{1}/{t}} \right]^{−\boldsymbol{{t}}\mathrm{cot}\:\mathrm{2}\boldsymbol{{x}}} \\ $$$$\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({e}^{−{t}\mathrm{cot}\:\mathrm{2}{x}} \right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{e}}^{−\boldsymbol{{y}}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{y}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[{t}\mathrm{cot}\:\mathrm{2}{x}=\left(\mathrm{2sin}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)\left(\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2sin}\:\frac{{x}}{\mathrm{3}}\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}\right)\right. \\ $$$${so}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{y}=\mathrm{0} \\ $$$${hence}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}\:{x}\right)^{\mathrm{cot}\:\mathrm{2}{x}} ={e}^{−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{y}} \\ $$$$\:\:=\mathrm{1}\:.\:\:\:\left({perhaps}\:{only}\:{right}\:{hand}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{llimit}\:\right)\:. \\ $$$$ \\ $$$$ \\ $$
Commented by tawa tawa last updated on 30/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/May/17
$${A}=\left({cosx}\right)^{{cot}\mathrm{2}{x}} \\ $$$${lnA}={cot}\mathrm{2}{x}.{lncosx} \\ $$$${lim}\left({l}\underset{{x}\rightarrow\mathrm{0}} {{n}A}\right)={l}\underset{{x}\rightarrow\mathrm{0}} {{i}m}\frac{{lncosx}}{{tgx}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{−{sinx}}{{cosx}}}{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}}= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(−{sinx}.{cosx}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{−\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{x}\right)=\mathrm{0} \\ $$$${lnA}=\mathrm{0}\Rightarrow{A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({cosx}\right)^{{cot}\mathrm{2}{x}} =\mathrm{1} \\ $$$${note}:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{tg}\mathrm{2}{x}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{tgx}=\mathrm{0} \\ $$$$ \\ $$
Commented by tawa tawa last updated on 30/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$