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Question Number 138888 by mathdanisur last updated on 19/Apr/21

$$Calculus:$$$$\underset{x\to \mathrm{\infty }}{lim}[\frac{x^2}{2^x}+\frac{2^x}{x!}+\frac{x!}{x^x}]=?$$

Answered by mathmax by abdo last updated on 20/Apr/21

$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\frac{{\mathrm{x}}^2}{2^{\mathrm{x}}}={\lim }_{\mathrm{x}\to +\mathrm{\infty }}{\mathrm{x}}^2{2}^{-\mathrm{x}}=0$$$$\mathrm{x}!\sim {\mathrm{x}}^{\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\sqrt{2\pi \mathrm{x}}(+\mathrm{\infty })\Rightarrow \frac{2^{\mathrm{x}}}{\mathrm{x}!}\sim \frac{2^{\mathrm{x}}}{{\mathrm{x}}^{\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\sqrt{2\pi \mathrm{x}}}=\frac{2^{\mathrm{x}}{\mathrm{x}}^{-\mathrm{x}}{\mathrm{e}}^{\mathrm{x}}}{\sqrt{2\pi \mathrm{x}}}$$$$=\frac{{\mathrm{e}}^{\mathrm{xlog}2}.{\mathrm{e}}^{-\mathrm{xlogx}}{\mathrm{e}}^{\mathrm{x}}}{\sqrt{2\pi \mathrm{x}}}=\frac{{\mathrm{e}}^{\mathrm{xlog}2-\mathrm{xlogx}+\mathrm{x}}}{\sqrt{2\pi \mathrm{x}}}=\frac{{\mathrm{e}}^{\mathrm{x}(\log 2-\mathrm{logx}+1)}}{\sqrt{2\pi \mathrm{x}}}\to 0(\mathrm{x}\to +\mathrm{\infty })$$$$\frac{\mathrm{x}!}{{\mathrm{x}}^{\mathrm{x}}}\sim {\mathrm{e}}^{-\mathrm{x}}\sqrt{2\pi \mathrm{x}}\to 0\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}(\frac{{\mathrm{x}}^2}{2^{\mathrm{x}}}+\frac{2^{\mathrm{x}}}{\mathrm{x}!}+\frac{\mathrm{x}!}{{\mathrm{x}}^{\mathrm{x}}})=0$$

Commented by mathdanisur last updated on 20/Apr/21

$$thankyouverymuchsir$$

Commented by Rasheed.Sindhi last updated on 20/Apr/21

$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}{\mathrm{x}}^2{2}^{-\mathrm{x}}=\mathrm{\infty }.0\left(indeterminateform\right)\ne 0$$$$Please\mathit{s}\mathit{i}\mathit{r}explain.$$

Commented by mnjuly1970 last updated on 20/Apr/21

$${lim}_{x\to \mathrm{\infty }}\frac{x^2}{2^x}\underset{hopital}{\stackrel{\frac{\mathrm{\infty }}{\mathrm{\infty }}}{=}}{lim}_{x\to +\mathrm{\infty }}\frac{2x}{2^x.ln\left(x\right)}$$$$={lim}_{x\to \mathrm{\infty }}\frac{2}{2^x.{ln}^2\left(x\right)}=\frac{\to 2}{\to \mathrm{\infty }}\to 0$$

Commented by Rasheed.Sindhi last updated on 20/Apr/21

$$\mathcal{B}utmathmaxsir{didn}^{\prime }tuse{l}^{\prime }hopital$$$$rule.Heevaluatedthelimitdirectly!$$$$\mathcal{A}nyway{you}^{\prime }rerivhtandthankssir!$$

Commented by Rasheed.Sindhi last updated on 20/Apr/21

$$\mathcal{T}h\alpha nksverymuchsir!$$

Commented by mathmax by abdo last updated on 20/Apr/21

$$\mathrm{i}\mathrm{use}\mathrm{that}{\mathrm{e}}^{\mathrm{x}}\mathrm{defeat}\mathrm{all}\mathrm{polynom}\mathrm{at}\mathrm{\infty }...!$$

Commented by mathmax by abdo last updated on 20/Apr/21

$$\mathrm{let}\mathrm{prove}\mathrm{this}\mathrm{result}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2{2}^{-\mathrm{x}}={\mathrm{x}}^2{\mathrm{e}}^{-\mathrm{xlog}2}$$$$\mathrm{changement}{\mathrm{e}}^{-\mathrm{xlog}2}=\mathrm{t}(\mathrm{t}\to 0)\Rightarrow -\mathrm{xlog}2=\mathrm{logt}\Rightarrow \mathrm{x}=-\frac{\mathrm{logt}}{\log 2}\Rightarrow$$$${\mathrm{x}}^2{\mathrm{e}}^{-\mathrm{xlog}2}=-\frac{{\log }^2\mathrm{t}}{{\log }^{2}2}\times \mathrm{t}=-\frac{{\mathrm{tlog}}^2\mathrm{t}}{{\log }^{2}2}\mathrm{due}\mathrm{to}{\lim }_{\mathrm{t}\to 0}\mathrm{tlogt}=0\Rightarrow$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}{\mathrm{x}}^2{\mathrm{e}}^{-\mathrm{xlog}2}=0$$$$$$

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