Show-why-lim-x-x-x-0-

Question Number 7472 by FilupSmith last updated on 31/Aug/16

$$\mathrm{Show}\:\mathrm{why}: \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{x}^{{x}} \:=\:\mathrm{0} \\ $$

Commented by FilupSmith last updated on 31/Aug/16

=lim_(x→−∞) e^(xln(x)) =lim_(x→−∞) e^(−∞∙ln(−∞)) ln(−∞)=ln(−1)+ln(∞)=iπ+∞ =lim_(x→−∞) e^(−∞∙(iπ+∞)) =lim_(x→−∞) e^(−∞i−∞) =lim_(x→−∞) e^(−(∞i+∞)) ????

$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{e}^{{x}\mathrm{ln}\left({x}\right)} \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{e}^{−\infty\centerdot\mathrm{ln}\left(−\infty\right)} \\ $$$$\mathrm{ln}\left(−\infty\right)=\mathrm{ln}\left(−\mathrm{1}\right)+\mathrm{ln}\left(\infty\right)={i}\pi+\infty \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{e}^{−\infty\centerdot\left({i}\pi+\infty\right)} \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{e}^{−\infty{i}−\infty} \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{e}^{−\left(\infty{i}+\infty\right)} \\ $$$$???? \\ $$

Answered by Yozzia last updated on 31/Aug/16

Write lim_(x→−∞) x^x =l. Let u=−x⇒ l=lim_(u→∞) (−u)^(−u) =lim_(u→∞) (1/((−u)^u )) l=lim_(u→∞) (1/((−1)^u u^u )). Because −1=e^(πi) and Euler′s formular gives e^(φi) =cosφ+isinφ, l=lim_(u→∞) (1/(e^(uπi) u^u ))=lim_(u→∞) (1/(u^u (cosπu+isinπu))) l=lim_(u→∞) ((cosπu−isinπu)/(u^u (cosπu+isinπu)(cosπu−isinπu)))=lim_(u→∞) ((cosπu−isinπu)/(u^u (cos^2 πu+sin^2 πu))) l=lim_(u→∞) ((cosπu−isinπu)/u^u ) since cos^2 πu+sin^2 πu=1. l=(lim_(u→∞) ((cosπu)/u^u ))−i(lim_(u→∞) ((sinπu)/u^u )) Since −1≤cosπu≤1⇒((−1)/u^u )≤((cosπu)/u^u )≤(1/u^u ) ⇒lim_(u→∞) ((−1)/u^u )≤lim_(u→∞) ((cosπu)/u^u )≤lim_(u→∞) (1/u^u ) ⇒0≤lim_(u→∞) ((cosπu)/u^u )≤0 Thus, lim_(u→∞) ((cosπu)/u^u )=0. Similarly, lim_(u→∞) ((sinπu)/u^u )=0. ∴l=0−i×0=0.

$${Write}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{x}^{{x}} ={l}. \\ $$$${Let}\:{u}=−{x}\Rightarrow\:{l}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\left(−{u}\right)^{−{u}} =\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\left(−{u}\right)^{{u}} } \\ $$$${l}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\left(−\mathrm{1}\right)^{{u}} {u}^{{u}} }. \\ $$$${Because}\:−\mathrm{1}={e}^{\pi{i}} \:{and}\:{Euler}'{s}\:{formular}\:{gives}\: \\ $$$${e}^{\phi{i}} ={cos}\phi+{isin}\phi, \\ $$$${l}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{e}^{{u}\pi{i}} {u}^{{u}} }=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{u}^{{u}} \left({cos}\pi{u}+{isin}\pi{u}\right)} \\ $$$${l}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{{cos}\pi{u}−{isin}\pi{u}}{{u}^{{u}} \left({cos}\pi{u}+{isin}\pi{u}\right)\left({cos}\pi{u}−{isin}\pi{u}\right)}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{{cos}\pi{u}−{isin}\pi{u}}{{u}^{{u}} \left({cos}^{\mathrm{2}} \pi{u}+{sin}^{\mathrm{2}} \pi{u}\right)} \\ $$$${l}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{{cos}\pi{u}−{isin}\pi{u}}{{u}^{{u}} }\:{since}\:{cos}^{\mathrm{2}} \pi{u}+{sin}^{\mathrm{2}} \pi{u}=\mathrm{1}. \\ $$$${l}=\left(\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{{cos}\pi{u}}{{u}^{{u}} }\right)−{i}\left(\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{{sin}\pi{u}}{{u}^{{u}} }\right) \\ $$$${Since}\:−\mathrm{1}\leqslant{cos}\pi{u}\leqslant\mathrm{1}\Rightarrow\frac{−\mathrm{1}}{{u}^{{u}} }\leqslant\frac{{cos}\pi{u}}{{u}^{{u}} }\leqslant\frac{\mathrm{1}}{{u}^{{u}} } \\ $$$$\Rightarrow\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{−\mathrm{1}}{{u}^{{u}} }\leqslant\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{{cos}\pi{u}}{{u}^{{u}} }\leqslant\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{u}^{{u}} } \\ $$$$\Rightarrow\mathrm{0}\leqslant\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{{cos}\pi{u}}{{u}^{{u}} }\leqslant\mathrm{0} \\ $$$${Thus},\:\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{{cos}\pi{u}}{{u}^{{u}} }=\mathrm{0}.\:{Similarly},\:\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{{sin}\pi{u}}{{u}^{{u}} }=\mathrm{0}. \\ $$$$\therefore{l}=\mathrm{0}−{i}×\mathrm{0}=\mathrm{0}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by FilupSmith last updated on 01/Sep/16

$$\mathrm{Amazing}! \\ $$

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