Question-112866

Question Number 112866 by bemath last updated on 10/Sep/20

Answered by bobhans last updated on 10/Sep/20

(7) L= lim_(x→0) ((1/x))^(tan x) L = e^(lim_(x→0) tan x.ln ((1/x))) =e^(lim_(x→0) −tan x.ln x) L = e^(lim_(x→0) ((−ln x)/(cot x))) = e^(lim_(x→0) ((−(1/x))/(−cosec^2 x))) L = e^(lim_(x→0) ((sin^2 x)/x)) = e^0 = 1

$$\left(\mathrm{7}\right)\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{tan}\:\mathrm{x}} \\ $$$$\:\:\:\mathrm{L}\:=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}tan}\:\mathrm{x}.\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)} =\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\mathrm{tan}\:\mathrm{x}.\mathrm{ln}\:\mathrm{x}} \\ $$$$\:\:\mathrm{L}\:=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{ln}\:\mathrm{x}}{\mathrm{cot}\:\mathrm{x}}} =\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\frac{\mathrm{1}}{\mathrm{x}}}{−\mathrm{cosec}\:^{\mathrm{2}} \mathrm{x}}} \\ $$$$\:\:\mathrm{L}\:=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{x}}} =\:\mathrm{e}^{\mathrm{0}} \:=\:\mathrm{1} \\ $$

Answered by Dwaipayan Shikari last updated on 10/Sep/20

5) lim_(x→0) ((e^x −e^(−x) )/(sinx))=((e^x −1)/x)+((1−(1/e^x ))/x)=1+((e^x −1)/(e^x x))=2

$$\left.\mathrm{5}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} −{e}^{−{x}} }{{sinx}}=\frac{{e}^{{x}} −\mathrm{1}}{{x}}+\frac{\mathrm{1}−\frac{\mathrm{1}}{{e}^{{x}} }}{{x}}=\mathrm{1}+\frac{{e}^{{x}} −\mathrm{1}}{{e}^{{x}} {x}}=\mathrm{2} \\ $$

Answered by Dwaipayan Shikari last updated on 10/Sep/20

lim_(x→0) ((e^x^2 −1)/(1−cosx))=lim_(x→0) ((e^x^2 −1)/(2sin^2 (x/2)))=2((e^x^2 −1)/x^2 )=2

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}^{\mathrm{2}} } −\mathrm{1}}{\mathrm{1}−{cosx}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}^{\mathrm{2}} } −\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}=\mathrm{2}\frac{{e}^{{x}^{\mathrm{2}} } −\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{2} \\ $$

Answered by mathmax by abdo last updated on 10/Sep/20

5) let use hospital theorem lim_(x→0) ((e^x −e^(−x) )/(sinx)) =lim_(x→0) ((2sh(x))/(sinx)) =lim_(x→0) ((2ch(x))/(cosx)) =2

$$\left.\mathrm{5}\right)\:\:\mathrm{let}\:\mathrm{use}\:\mathrm{hospital}\:\mathrm{theorem}\:\: \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{sinx}}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{\mathrm{2sh}\left(\mathrm{x}\right)}{\mathrm{sinx}}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{2ch}\left(\mathrm{x}\right)}{\mathrm{cosx}}\:=\mathrm{2} \\ $$

Answered by mathmax by abdo last updated on 10/Sep/20

6) let f(x) =((e^x^2 −1)/(1−cosx)) we have e^u ∼1+u ⇒e^x^2 ∼1+x^2 and 1−cosx ∼(x^2 /2) ⇒f(x) ∼(x^2 /(x^2 /2))=2 ⇒lim_(x→0) f(x) =0

$$\left.\mathrm{6}\right)\:\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } −\mathrm{1}}{\mathrm{1}−\mathrm{cosx}}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{e}^{\mathrm{u}} \:\sim\mathrm{1}+\mathrm{u}\:\Rightarrow\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \:\sim\mathrm{1}+\mathrm{x}^{\mathrm{2}} \:\mathrm{and} \\ $$$$\mathrm{1}−\mathrm{cosx}\:\sim\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\sim\frac{\mathrm{x}^{\mathrm{2}} }{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}}=\mathrm{2}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{0} \\ $$

Answered by mathmax by abdo last updated on 10/Sep/20

7) let g(x)=((1/x))^(tanx) ⇒g(x) =e^(tanxln((1/x))) =e^(−ln(x)tanx) ⇒g(x)∼e^(−xlnx) ⇒lim_(x→0) g(x) =e^0 =1

$$\left.\mathrm{7}\right)\:\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)=\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{tanx}} \:\Rightarrow\mathrm{g}\left(\mathrm{x}\right)\:=\mathrm{e}^{\mathrm{tanxln}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)} \:=\mathrm{e}^{−\mathrm{ln}\left(\mathrm{x}\right)\mathrm{tanx}} \\ $$$$\Rightarrow\mathrm{g}\left(\mathrm{x}\right)\sim\mathrm{e}^{−\mathrm{xlnx}} \:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\mathrm{g}\left(\mathrm{x}\right)\:=\mathrm{e}^{\mathrm{0}} \:=\mathrm{1} \\ $$

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