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Question Number 195653 by mustafazaheen last updated on 06/Aug/23

_( →0) ( 1)^((√3)/ ) =?

$$\underset{ \rightarrow\mathrm{0}} { }\left( \mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{ }} \:=? \\ $$$$ \\ $$

Answered by MM42 last updated on 06/Aug/23

e−1>1 ⇒lim_(x→ 0^+ ) (e−1)^((√3)/x) =∞ ⇒lim_(x→ 0^− ) (e−1)^((√3)/x) =0 ⇒lim no exist

$${e}−\mathrm{1}>\mathrm{1}\:\:\: \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\:\mathrm{0}^{+} } \:\left({e}−\mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{{x}}} =\infty \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\:\mathrm{0}^{−} } \:\left({e}−\mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{{x}}} =\mathrm{0} \\ $$$$\Rightarrow{lim}\:\:{no}\:\:{exist} \\ $$$$ \\ $$

Answered by mr W last updated on 06/Aug/23

e−1>1 lim_(n→+∞) (e−1)^n =+∞ lim_(n→−∞) (e−1)^n =0 lim_(x→0^+ ) (e−1)^((√3)/x) =+∞ lim_(x→0^− ) (e−1)^((√3)/x) =0 ⇒lim_(x→0) (e−1)^((√3)/x) doesn′t exist!

$${e}−\mathrm{1}>\mathrm{1} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\left({e}−\mathrm{1}\right)^{{n}} =+\infty \\ $$$$\underset{{n}\rightarrow−\infty} {\mathrm{lim}}\left({e}−\mathrm{1}\right)^{{n}} =\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left({e}−\mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{{x}}} =+\infty \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\left({e}−\mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{{x}}} =\mathrm{0} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({e}−\mathrm{1}\right)^{\frac{\sqrt{\mathrm{3}}}{{x}}} \:{doesn}'{t}\:{exist}! \\ $$

Commented by mustafazaheen last updated on 07/Aug/23

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