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lim-x-0-cot-x-1-ln-x-




Question Number 121460 by bramlexs22 last updated on 08/Nov/20
 lim_(x→0^+ )  (cot x)^(1/(ln x))  ?
$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left(\mathrm{cot}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{x}}} \:?\: \\ $$
Answered by Dwaipayan Shikari last updated on 08/Nov/20
lim_(x→0) (cotx)^(1/(logx)) =y  (1/(logx))log(((cosx)/(sinx)))=logy         (cosx→1)  ⇒−((log(sinx))/(logx))=logy  =−1=logy⇒y=(1/e)  (sinx→x)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({cotx}\right)^{\frac{\mathrm{1}}{{logx}}} ={y} \\ $$$$\frac{\mathrm{1}}{{logx}}{log}\left(\frac{{cosx}}{{sinx}}\right)={logy}\:\:\:\:\:\:\:\:\:\left({cosx}\rightarrow\mathrm{1}\right) \\ $$$$\Rightarrow−\frac{{log}\left({sinx}\right)}{{logx}}={logy} \\ $$$$=−\mathrm{1}={logy}\Rightarrow{y}=\frac{\mathrm{1}}{{e}}\:\:\left({sinx}\rightarrow{x}\right) \\ $$
Answered by liberty last updated on 08/Nov/20
 L= lim_(x→0^+ )  (cot x)^(1/(ln x))    ln L = lim_(x→0^+ )  ((ln cot x)/(ln x)) = lim_(x→0^+ )  ((−cosec^2 x)/((1/x).cot x))   ln L = −lim_(x→0^+ )  ((xsin x)/(sin^2 x.cos x)) = −1   L = e^(−1)  = (1/e).
$$\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left(\mathrm{cot}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{x}}} \\ $$$$\:\mathrm{ln}\:\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{ln}\:\mathrm{cot}\:\mathrm{x}}{\mathrm{ln}\:\mathrm{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{−\mathrm{cosec}\:^{\mathrm{2}} \mathrm{x}}{\frac{\mathrm{1}}{\mathrm{x}}.\mathrm{cot}\:\mathrm{x}} \\ $$$$\:\mathrm{ln}\:\mathrm{L}\:=\:−\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{xsin}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}.\mathrm{cos}\:\mathrm{x}}\:=\:−\mathrm{1} \\ $$$$\:\mathrm{L}\:=\:\mathrm{e}^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{e}}. \\ $$

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