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Question Number 136701 by mathlove last updated on 25/Mar/21

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{3^n-n!}{2^n-n!}=?$$

Commented by yutytfjh67ihd last updated on 25/Mar/21

Commented by mathlove last updated on 25/Mar/21

$$helpme$$

Answered by mathmax by abdo last updated on 25/Mar/21

$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{\mathrm{n}!-3^{\mathrm{n}}}{\mathrm{n}!-2^{\mathrm{n}}}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{1-\frac{3^{\mathrm{n}}}{\mathrm{n}!}}{1-\frac{2^{\mathrm{n}}}{\mathrm{n}!}}$$$$\mathrm{n}!\sim {\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}\Rightarrow \frac{3^{\mathrm{n}}}{\mathrm{n}!}\sim \frac{3^{\mathrm{n}}}{{\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}}\Rightarrow$$$$\log \left(\frac{3^{\mathrm{n}}}{\mathrm{n}!}\right)\sim \mathrm{nlog}3-\mathrm{nlogn}+\mathrm{n}-\frac{1}{2}\log \left(2\mathrm{n}\pi \right)$$$$=\mathrm{n}(\log 3-\mathrm{logn}+1-\frac{1}{2}\log \left(2\mathrm{n}\pi \right))\to -\mathrm{\infty }\Rightarrow \frac{3^{\mathrm{n}}}{\mathrm{n}!}\to 0$$$$\mathrm{also}\frac{2^{\mathrm{n}}}{\mathrm{n}!}\to 0\Rightarrow {\lim }_{\mathrm{n}\to \mathrm{\infty }}\frac{\mathrm{n}!-3^{\mathrm{n}}}{\mathrm{n}!-2^{\mathrm{n}}}=1$$

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