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Question Number 92423 by mhmd last updated on 06/May/20

$${lim}_{x\Rightarrow \mathrm{\infty }}\frac{4(x+3)!-x!}{x[(x+2)!-(x-1)!]}$$

Answered by john santu last updated on 07/May/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{4(\mathrm{x}+3)(\mathrm{x}+2)(\mathrm{x}+1)\mathrm{x}!-\mathrm{x}!}{\mathrm{x}(\mathrm{x}+2)(\mathrm{x}+1)\mathrm{x}!-\mathrm{x}(\mathrm{x}-1)!}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\mathrm{x}!(4(\mathrm{x}+3)(\mathrm{x}+2)(\mathrm{x}+1)-1)}{\mathrm{x}!(\mathrm{x}(\mathrm{x}+2)(\mathrm{x}+1)-1)}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{4(\mathrm{x}+3)(\mathrm{x}+2)(\mathrm{x}+1)-1}{\mathrm{x}(\mathrm{x}+2)(\mathrm{x}+1)-1}$$$$=4$$

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