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$\lim _{x \rightarrow \infty}\left(\frac{x+2}{x+1}\right)^{x+3}$
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Verified Answer
The correct answer is:
$e$
Let $A=\lim _{x \rightarrow \infty}\left(\frac{x+2}{x+1}\right)^{x+3}$
$\begin{array}{l}
=\lim _{x \rightarrow \infty}\left(1+\frac{1}{x+1}\right)^{x+3}=\lim _{x \rightarrow \infty}\left[\left(1+\frac{1}{x+1}\right)^{x+1}\right]^{\frac{(x+3)}{(x+1)}}=e \\
\left\{\because \lim _{x \rightarrow \infty}\left(1+\frac{1}{x+1}\right)^{x+1}=e\right. \text { and } \left.\lim _{x \rightarrow \infty} \frac{(x+3)}{(x+1)}=\lim _{x \rightarrow \infty} \frac{\{1+(3 / x)\}}{\{1+(1 / x)\}}=1\right\} .
\end{array}$
$\begin{array}{l}
=\lim _{x \rightarrow \infty}\left(1+\frac{1}{x+1}\right)^{x+3}=\lim _{x \rightarrow \infty}\left[\left(1+\frac{1}{x+1}\right)^{x+1}\right]^{\frac{(x+3)}{(x+1)}}=e \\
\left\{\because \lim _{x \rightarrow \infty}\left(1+\frac{1}{x+1}\right)^{x+1}=e\right. \text { and } \left.\lim _{x \rightarrow \infty} \frac{(x+3)}{(x+1)}=\lim _{x \rightarrow \infty} \frac{\{1+(3 / x)\}}{\{1+(1 / x)\}}=1\right\} .
\end{array}$
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