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Question Number 168014 by mathlove last updated on 31/Mar/22

lim_(x→0) ((5e^x +5e^(−x) −10)/(4x^2 ))=?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{5}{e}^{{x}} +\mathrm{5}{e}^{−{x}} −\mathrm{10}}{\mathrm{4}{x}^{\mathrm{2}} }=? \\ $$

Answered by som(math1967) last updated on 01/Apr/22

((5e^x +(5/e^x )−10)/(4x^2 )) =((5e^x −5−(5/e^x )(e^x −1))/(4x^2 )) =((5(e^x −1)−(5/e^x )(e^x −1))/(4x^2 )) =(((e^x −1)(5−(5/e^x )))/(4x^2 )) =((5(e^x −1)(e^x −1))/(4x^2 e^x )) ∴lim_(x→0) (5/4)×(((e^x −1)/x))^2 ×(1/e^x ) =(5/4)×1^2 ×1=(5/4)

$$\:\:\frac{\mathrm{5}{e}^{{x}} +\frac{\mathrm{5}}{{e}^{{x}} }−\mathrm{10}}{\mathrm{4}{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{5}{e}^{{x}} −\mathrm{5}−\frac{\mathrm{5}}{{e}^{{x}} }\left({e}^{{x}} −\mathrm{1}\right)}{\mathrm{4}{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{5}\left({e}^{{x}} −\mathrm{1}\right)−\frac{\mathrm{5}}{{e}^{{x}} }\left({e}^{{x}} −\mathrm{1}\right)}{\mathrm{4}{x}^{\mathrm{2}} } \\ $$$$=\frac{\left({e}^{{x}} −\mathrm{1}\right)\left(\mathrm{5}−\frac{\mathrm{5}}{{e}^{{x}} }\right)}{\mathrm{4}{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{5}\left({e}^{{x}} −\mathrm{1}\right)\left({e}^{{x}} −\mathrm{1}\right)}{\mathrm{4}{x}^{\mathrm{2}} {e}^{{x}} } \\ $$$$\therefore\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{\mathrm{5}}{\mathrm{4}}×\left(\frac{{e}^{{x}} −\mathrm{1}}{{x}}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{{e}^{{x}} } \\ $$$$=\frac{\mathrm{5}}{\mathrm{4}}×\mathrm{1}^{\mathrm{2}} ×\mathrm{1}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$

Commented by som(math1967) last updated on 01/Apr/22

other way (5/4)lim_(x→0) ((e^x +e^(−x) −2)/x^2 ) =(5/4)lim_(x→0) ((e^(2x) +1−2e^x )/(e^x ×x^2 )) =(5/4) lim_(x→0) (((e^x −1)/x))^2 ×(1/e^x ) =(5/4)×1^2 ×1=(5/4)

$$\boldsymbol{{other}}\:\boldsymbol{{way}} \\ $$$$\frac{\mathrm{5}}{\mathrm{4}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{{e}^{{x}} +{e}^{−{x}} −\mathrm{2}}{{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{5}}{\mathrm{4}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}} +\mathrm{1}−\mathrm{2}\boldsymbol{{e}}^{\boldsymbol{{x}}} }{\boldsymbol{{e}}^{\boldsymbol{{x}}} ×\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{5}}{\mathrm{4}}\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\left(\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}}{\boldsymbol{{x}}}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\boldsymbol{{e}}^{\boldsymbol{{x}}} } \\ $$$$=\frac{\mathrm{5}}{\mathrm{4}}×\mathrm{1}^{\mathrm{2}} ×\mathrm{1}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$ \\ $$

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