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lim-x-0-cosh-2x-cosh-x-1-x-2-




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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