Limits Questions

Question Number 163657 by mathlove last updated on 09/Jan/22

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} −{e}}{{x}}=? \\$$

Answered by mr W last updated on 09/Jan/22

$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({e}^{\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}}} \right)' \\$$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}−\frac{\mathrm{1}}{{x}}\mathrm{ln}\:\left(\mathrm{1}+{x}\right)\right)\frac{\mathrm{1}}{{x}} \\$$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \left(\mathrm{1}−{x}+{x}^{\mathrm{2}} −...−\frac{\left({x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−...\right)}{{x}}\right)\frac{\mathrm{1}}{{x}} \\$$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \left(\mathrm{1}−{x}+{x}^{\mathrm{2}} −...−\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−...\right)\right)\frac{\mathrm{1}}{{x}} \\$$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \left(−\frac{{x}}{\mathrm{2}}+\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{3}}−....\right)\frac{\mathrm{1}}{{x}} \\$$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}{x}}{\mathrm{3}}−....\right) \\$$$$={e}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\$$$$=−\frac{{e}}{\mathrm{2}} \\$$

Commented by mathlove last updated on 09/Jan/22

$${thanks}\:\:{mr}\:\:{W} \\$$

Answered by mnjuly1970 last updated on 09/Jan/22

$$\:\:\:{lim}_{\:{x}\rightarrow\mathrm{0}} \:\frac{\:{e}^{\:\frac{\mathrm{1}}{{x}}\:{ln}\left(\mathrm{1}+{x}\right)} −{e}}{{x}} \\$$$$\:\:={lim}_{\:{x}\rightarrow\mathrm{0}} \frac{\:{e}^{\:\frac{\mathrm{1}}{{x}}\:\left(\:{x}−\frac{{x}^{\:\mathrm{2}} }{\mathrm{2}}\:+\:{O}\left({x}^{\:\mathrm{2}} \right)\right)} −{e}}{{x}} \\$$$$\:=\:{elim}_{\:\:{x}\rightarrow\mathrm{0}} \frac{\:−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{\:−{x}} }{\mathrm{1}}\:=\frac{−{e}}{\mathrm{2}} \\$$

Commented by mathlove last updated on 10/Jan/22

$${thanks} \\$$