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Question Number 77442 by aliesam last updated on 06/Jan/20

$$provethat$$$$\underset{x\to \mathrm{\infty }}{lim}(\mid \frac{x^{x^2}{(x+2)}^{{(x+1)}^2}}{{(x+1)}^{2x^2+2x+1}}\mid )=e$$

Answered by aliesam last updated on 06/Jan/20

$$L=\underset{x\to \mathrm{\infty }}{lim}\mid \left(\frac{x^{x^2}}{{(x+1)}^{x^2}}\right)\times \left(\frac{{(x+2)}^{{(x+1)}^2}}{{(x+1)}^{{(x+1)}^2}}\right)\mid$$$$L=\underset{x\to \mathrm{\infty }}{lim}(\mid {\left(\frac{x}{x+1}\right)}^{x^2}\times {\left(\frac{x+2}{x+1}\right)}^{{(x+1)}^2}\mid$$$$L=\underset{x\to \mathrm{\infty }}{lim}\mid {(1-\frac{1}{x+1})}^{x^2}\times {(1+\frac{1}{x+1})}^{{(x+1)}^2}\mid$$$$L=\underset{x\to \mathrm{\infty }}{lim}{e}^{\mid x^{2}ln(1-\frac{1}{x+1})+{(x+1)}^{2}ln(1+\frac{1}{x+1})\mid }$$$$\underset{z\to 0}{lim}L=\underset{x\to \mathrm{\infty }}{limL}whenz=\frac{1}{x+1}and{x}^2=\frac{1}{z^2}-\frac{2}{z}+1$$$${(x+1)}^2=\frac{1}{z^2}$$$$$$$$L=\underset{z\to 0}{lim}{e}^{\mid \frac{ln(1-z)}{z^2}-\frac{2ln(1-z)}{z}+ln(1-z)+\frac{ln(1+z)}{z^2}\mid }$$$$L=e^{-1+2}=e$$

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