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Question Number 117637 by bemath last updated on 13/Oct/20

lim_(x→0)  (cos x )^(cot x)  =?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}\:\mathrm{x}\:\right)^{\mathrm{cot}\:\mathrm{x}} \:=? \\ $$

Answered by AbduraufKodiriy last updated on 13/Oct/20

lim_(x→0) (cosx^(cotx) )=lim_(x→0) (cosx^((cosx)/(sinx)) )=  =lim_(x→0) (1+cosx−1)^((cosxsinx)/(1−cos^2 x)) =lim_(x→0) ((1+cosx−1)^(1/(1−cosx)) )^((cosxsinx)/(1+cosx)) =  =lim_(x→0) (e^((cosxsinx)/(1+cosx)) )=e^0 =1

$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\left(\boldsymbol{{cosx}}^{\boldsymbol{{cotx}}} \right)=\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\left(\boldsymbol{{cosx}}^{\frac{\boldsymbol{{cosx}}}{\boldsymbol{{sinx}}}} \right)= \\ $$$$=\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\left(\mathrm{1}+\boldsymbol{{cosx}}−\mathrm{1}\right)^{\frac{\boldsymbol{{cosxsinx}}}{\mathrm{1}−\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}}} =\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\left(\left(\mathrm{1}+\boldsymbol{{cosx}}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{{cosx}}}} \right)^{\frac{\boldsymbol{{cosxsinx}}}{\mathrm{1}+\boldsymbol{{cosx}}}} = \\ $$$$=\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\left(\boldsymbol{{e}}^{\frac{\boldsymbol{{cosxsinx}}}{\mathrm{1}+\boldsymbol{{cosx}}}} \right)=\boldsymbol{{e}}^{\mathrm{0}} =\mathrm{1} \\ $$

Answered by Dwaipayan Shikari last updated on 13/Oct/20

lim_(x→0) (cosx)^(cotx) =y  cotxlog(1+cosx−1)=logy  cotx(cosx−1)=logy               lim_(x→0) log(1+x)=x  ((cos^2 x−cosx)/(sinx))=logy  cosx(((2sin^2 (x/2))/(2sin(x/2)cos(x/2))))=logy  logy=0  y=e^0 =1

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({cosx}\right)^{{cotx}} ={y} \\ $$$${cotxlog}\left(\mathrm{1}+{cosx}−\mathrm{1}\right)={logy} \\ $$$${cotx}\left({cosx}−\mathrm{1}\right)={logy}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{log}\left(\mathrm{1}+{x}\right)={x} \\ $$$$\frac{{cos}^{\mathrm{2}} {x}−{cosx}}{{sinx}}={logy} \\ $$$${cosx}\left(\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}\right)={logy} \\ $$$${logy}=\mathrm{0} \\ $$$${y}={e}^{\mathrm{0}} =\mathrm{1} \\ $$$$ \\ $$

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