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Question Number 148416 by mathdanisur last updated on 27/Jul/21

$$\underset{\mathit{x}\to \mathrm{\infty }}{lim}\frac{x!-cos\left(2x\right)}{3x+1}=?$$

Answered by mathmax by abdo last updated on 28/Jul/21

$$\mathrm{we}\mathrm{have}\frac{\mathrm{x}!-\cos \left(2\mathrm{x}\right)}{3\mathrm{x}+1}=\frac{\mathrm{x}!}{3\mathrm{x}+1}-\frac{\cos \left(2\mathrm{x}\right)}{3\mathrm{x}+1}$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{3\mathrm{x}+1}=0\mathrm{because}\mid \cos \left(2\mathrm{x}\right)\mid ⩽1$$$$\mathrm{x}!\sim {\mathrm{x}}^{\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\sqrt{2\pi \mathrm{x}}\Rightarrow \frac{\mathrm{x}!}{3\mathrm{x}+1}\sim \frac{1}{3}\times \frac{\mathrm{x}!}{\mathrm{x}}\sim \frac{1}{3}{\mathrm{x}}^{\mathrm{x}-1}{\mathrm{e}}^{-\mathrm{x}}\sqrt{2\pi \mathrm{x}}$$$$=\frac{1}{3}{\mathrm{e}}^{(\mathrm{x}-1)\mathrm{logx}-\mathrm{x}}\sqrt{2\pi \mathrm{x}}=\frac{1}{3}\sqrt{2\pi }{\mathrm{e}}^{(\mathrm{x}-1)\mathrm{logx}-\frac{1}{2}}\to +\mathrm{\infty }(\mathrm{x}\to +\mathrm{\infty })\Rightarrow$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\frac{\mathrm{x}!-\cos \left(2\mathrm{x}\right)}{3\mathrm{x}+1}=+\mathrm{\infty }$$

Commented by mathdanisur last updated on 28/Jul/21

$$ThankyoySer$$

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