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Question Number 117147 by bemath last updated on 10/Oct/20
lim_(x→0) (((cosh (2x))/(cosh (x))))^(1/x^2 ) =?
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{cosh}\:\left(\mathrm{2x}\right)}{\mathrm{cosh}\:\left(\mathrm{x}\right)}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \:=? \\ $$
Answered by Olaf last updated on 10/Oct/20
coshu ∼_0 1+(u^2 /2) ((cosh2x)/(coshx)) ∼_0 ((1+2x^2 )/(1+(x^2 /2))) ∼_0 (1+2x^2 )(1−(x^2 /2)) ((cosh2x)/(coshx)) ∼_0 1+(3/2)x^2 ln(((cosh2x)/(coshx))) ∼_0 ln(1+(3/2)x^2 ) ∼_0 (3/2)x^2 (1/x^2 )ln(((cosh2x)/(coshx))) ∼_0 (3/2) ln(((cosh2x)/(coshx)))^(1/x^2 ) ∼_0 (3/2) (((cosh2x)/(coshx)))^(1/x^2 ) ∼_0 e^(3/2)
$$\mathrm{cosh}{u}\:\underset{\mathrm{0}} {\sim}\:\mathrm{1}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{\mathrm{cosh2}{x}}{\mathrm{cosh}{x}}\:\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}\:\underset{\mathrm{0}} {\sim}\:\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{cosh2}{x}}{\mathrm{cosh}{x}}\:\underset{\mathrm{0}} {\sim}\:\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \\ $$$$\mathrm{ln}\left(\frac{\mathrm{cosh2}{x}}{\mathrm{cosh}{x}}\right)\:\underset{\mathrm{0}} {\sim}\:\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \right)\:\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\mathrm{ln}\left(\frac{\mathrm{cosh2}{x}}{\mathrm{cosh}{x}}\right)\:\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{ln}\left(\frac{\mathrm{cosh2}{x}}{\mathrm{cosh}{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{cosh2}{x}}{\mathrm{cosh}{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:\underset{\mathrm{0}} {\sim}\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by bemath last updated on 10/Oct/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by bobhans last updated on 10/Oct/20
L= lim_(x→0) (((cosh (2x))/(cosh (x))))^(1/x^2 ) = lim_(x→0) ((((1/2)(e^(2x) +e^(−2x) ))/((1/2)(e^x +e^(−x) ))))^(1/x^2 ) ln L= lim_(x→0) ((ln (e^(2x) +e^(−2x) )−ln (e^x +e^(−x) ))/x^2 ) ln L = lim_(x→0) ((((2e^(2x) −2e^(−2x) )/(e^(2x) +e^(−2x) )) −((e^x −e^(−x) )/(e^x +e^(−x) )))/(2x)) ln L = (1/2)lim_(x→0) ((2e^(2x) −2e^(−2x) −e^x +e^(−x) )/(2x)) ln L = (1/2)lim_(x→0) ((4e^(2x) +4e^(−2x) −e^x −e^(−x) )/2) L = e^((1/2).3) = e^(3/2) = (√e^3 )
$$\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{cosh}\:\left(\mathrm{2x}\right)}{\mathrm{cosh}\:\left(\mathrm{x}\right)}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} =\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{2x}} +\mathrm{e}^{−\mathrm{2x}} \right)}{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} \right)}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{ln}\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{e}^{\mathrm{2x}} +\mathrm{e}^{−\mathrm{2x}} \right)−\mathrm{ln}\:\left(\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} \right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2e}^{\mathrm{2x}} −\mathrm{2e}^{−\mathrm{2x}} }{\mathrm{e}^{\mathrm{2x}} +\mathrm{e}^{−\mathrm{2x}} }\:−\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }}{\mathrm{2x}} \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2e}^{\mathrm{2x}} −\mathrm{2e}^{−\mathrm{2x}} −\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{2x}} \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4e}^{\mathrm{2x}} +\mathrm{4e}^{−\mathrm{2x}} −\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}} \\ $$$$\mathrm{L}\:=\:\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{3}} =\:\mathrm{e}^{\frac{\mathrm{3}}{\mathrm{2}}} \:=\:\sqrt{\mathrm{e}^{\mathrm{3}} }\: \\ $$

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