Question Number 194344 by mathlove last updated on 04/Jul/23
Answered by qaz last updated on 04/Jul/23
$$\mathrm{sin}\:^{\mathrm{2}} {x}=\left({x}−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +…\right)^{\mathrm{2}} ={x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{4}} +… \\ $$$${e}^{{x}} +{e}^{−{x}} −{x}^{\mathrm{2}} −\mathrm{2}=\frac{\mathrm{1}}{\mathrm{24}}{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{24}}\left(−{x}\right)^{\mathrm{4}} +…=\frac{\mathrm{1}}{\mathrm{12}}{x}^{\mathrm{4}} +… \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{e}^{{x}} +{e}^{−{x}} −{x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{sin}\:^{\mathrm{2}} {x}−{x}^{\mathrm{2}} }=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\frac{\mathrm{1}}{\mathrm{12}}{x}^{\mathrm{4}} }{−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{4}} }=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Answered by anurup last updated on 04/Jul/23
![lim_(x→0) ((e^x −e^(−x) −2x)/( 2sin xcos x−2x)) [by L′Hospital′s Rule] =lim_(x→0) ((e^x −e^(−x) −2x)/(sin 2x−2x)) =lim_(x→0) ((e^x +e^(−x) −2)/(2cos 2x−2)) [by L′Hospital′s rule] =lim_(x→0) ((e^x −e^(−x) )/(−4sin 2x)) [by L′Hospital′s rule] =lim_(x→0) ((e^x +e^(−x) )/(−8cos 2x)) [by L′Hospital′s rule] =−(1/4)](https://www.tinkutara.com/question/Q194349.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} −{e}^{−{x}} −\mathrm{2}{x}}{\:\:\mathrm{2sin}\:{x}\mathrm{cos}\:{x}−\mathrm{2}{x}}\:\left[\mathrm{by}\:\mathrm{L}'\mathrm{Hospital}'\mathrm{s}\:\mathrm{Rule}\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{{x}} −{e}^{−{x}} −\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} +{e}^{−{x}} −\mathrm{2}}{\mathrm{2cos}\:\mathrm{2}{x}−\mathrm{2}}\:\left[\mathrm{by}\:\mathrm{L}'\mathrm{Hospital}'\mathrm{s}\:\mathrm{rule}\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −{e}^{−{x}} }{−\mathrm{4sin}\:\mathrm{2}{x}}\:\:\left[\mathrm{by}\:\mathrm{L}'\mathrm{Hospital}'\mathrm{s}\:\mathrm{rule}\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} +{e}^{−{x}} }{−\mathrm{8cos}\:\mathrm{2}{x}}\:\left[\mathrm{by}\:\mathrm{L}'\mathrm{Hospital}'\mathrm{s}\:\mathrm{rule}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$