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Question Number 6407 by FilupSmith last updated on 26/Jun/16

Can you evaluate: lim_(x→∞) (−1)^x x WolframAlpha writes the answer as =e^(2i to π) ∞

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{evaluate}: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(−\mathrm{1}\right)^{{x}} {x} \\ $$$$ \\ $$$$\mathrm{WolframAlpha}\:\mathrm{writes}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{as} \\ $$$$={e}^{\mathrm{2}{i}\:\:\mathrm{to}\:\:\pi} \infty \\ $$

Commented by FilupSmith last updated on 26/Jun/16

Is there a more formal way to write the answer?

$$\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{more}\:\mathrm{formal}\:\mathrm{way}\:\mathrm{to}\:\mathrm{write} \\ $$$$\mathrm{the}\:\mathrm{answer}? \\ $$

Commented by FilupSmith last updated on 26/Jun/16

I know that: (−1)^x =e^(xln(−1)) =e^(iπx) e^(iπx) =cos(πx)+isin(πx) −1≤cos(πx)≤1 −i≤isin(πx)≤i So is (−1)^x , x∈R −(1+i)≤(−1)^x ≤1+i ???

$$\mathrm{I}\:\mathrm{know}\:\mathrm{that}: \\ $$$$\left(−\mathrm{1}\right)^{{x}} ={e}^{{x}\mathrm{ln}\left(−\mathrm{1}\right)} ={e}^{{i}\pi{x}} \\ $$$${e}^{{i}\pi{x}} =\mathrm{cos}\left(\pi{x}\right)+{i}\mathrm{sin}\left(\pi{x}\right) \\ $$$$−\mathrm{1}\leqslant\mathrm{cos}\left(\pi{x}\right)\leqslant\mathrm{1} \\ $$$$−{i}\leqslant{i}\mathrm{sin}\left(\pi{x}\right)\leqslant{i} \\ $$$$ \\ $$$$\mathrm{So}\:\mathrm{is}\:\left(−\mathrm{1}\right)^{{x}} ,\:{x}\in\mathbb{R} \\ $$$$−\left(\mathrm{1}+{i}\right)\leqslant\left(−\mathrm{1}\right)^{{x}} \leqslant\mathrm{1}+{i} \\ $$$$??? \\ $$

Answered by Temp last updated on 27/Jun/16

(−1)^x =e^(iπx) e^(iπx) =cos(xπ)+isin(xπ) ℜ(e^(iπx) )=cos(πx) −1≤ℜ(e^(iπx) )≤1 ℑ(e^(iπx) )=sin(πx) −1≤ℑ(e^(iπx) )≤1 Let L=lim_(x→∞) (e^(iπx) ) if L∈R⇒x∈Z:−1≤x≤1 ∴(−1)^x = { ((1 if 2∣x)),((−1 if 2∤x)) :} if L∈C⇒x∈R −(i+1)≤L≤(i+1)

$$\left(−\mathrm{1}\right)^{{x}} ={e}^{{i}\pi{x}} \\ $$$${e}^{{i}\pi{x}} =\mathrm{cos}\left({x}\pi\right)+{i}\mathrm{sin}\left({x}\pi\right) \\ $$$$ \\ $$$$\Re\left({e}^{{i}\pi{x}} \right)=\mathrm{cos}\left(\pi{x}\right) \\ $$$$−\mathrm{1}\leqslant\Re\left({e}^{{i}\pi{x}} \right)\leqslant\mathrm{1} \\ $$$$ \\ $$$$\Im\left({e}^{{i}\pi{x}} \right)=\mathrm{sin}\left(\pi{x}\right) \\ $$$$−\mathrm{1}\leqslant\Im\left({e}^{{i}\pi{x}} \right)\leqslant\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Let}\:{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({e}^{{i}\pi{x}} \right) \\ $$$$\mathrm{if}\:{L}\in\mathbb{R}\Rightarrow{x}\in\mathbb{Z}:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\therefore\left(−\mathrm{1}\right)^{{x}} =\begin{cases}{\mathrm{1}\:\:\:\mathrm{if}\:\mathrm{2}\mid{x}}\\{−\mathrm{1}\:\:\mathrm{if}\:\mathrm{2}\nmid{x}}\end{cases} \\ $$$$\mathrm{if}\:{L}\in\mathbb{C}\Rightarrow{x}\in\mathbb{R} \\ $$$$−\left({i}+\mathrm{1}\right)\leqslant{L}\leqslant\left({i}+\mathrm{1}\right) \\ $$

Commented by Temp last updated on 27/Jun/16

Im(x)=ℑ(x) Re(x)=ℜ(x)

$$\mathrm{Im}\left({x}\right)=\Im\left({x}\right) \\ $$$$\mathrm{Re}\left({x}\right)=\Re\left({x}\right) \\ $$

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