Question Number 195653 by mustafazaheen last updated on 06/Aug/23
Answered by MM42 last updated on 06/Aug/23
$${e}−\mathrm{1}>\mathrm{1}\:\:\: \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\:\mathrm{0}^{+} } \:\left({e}−\mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{{x}}} =\infty \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\:\mathrm{0}^{−} } \:\left({e}−\mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{{x}}} =\mathrm{0} \\ $$$$\Rightarrow{lim}\:\:{no}\:\:{exist} \\ $$$$ \\ $$
Answered by mr W last updated on 06/Aug/23
$${e}−\mathrm{1}>\mathrm{1} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\left({e}−\mathrm{1}\right)^{{n}} =+\infty \\ $$$$\underset{{n}\rightarrow−\infty} {\mathrm{lim}}\left({e}−\mathrm{1}\right)^{{n}} =\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left({e}−\mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{{x}}} =+\infty \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left({e}−\mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{{x}}} =\mathrm{0} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({e}−\mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{{x}}} \:{doesn}'{t}\:{exist}! \\ $$
Commented by mustafazaheen last updated on 07/Aug/23
