问题及解答

\[
\lim_{x\rightarrow -\infty}\biggl(\frac{x+1}{x-2}\biggr)^{2x-1},
\]

\[
\lim_{x\rightarrow -\infty}e^{(\frac{x+1}{x-2})^{2x-1}}.
\]

\[
\begin{split}
\lim_{x\rightarrow -\infty}\biggl(\frac{x+1}{x-2}\biggr)^{2x-1}&=\lim_{x\rightarrow -\infty}\biggl(1+\frac{3}{x-2}\biggr)^{2x-1}\\
&=\lim_{x\rightarrow -\infty}\biggl(1+\frac{1}{\frac{x-2}{3}}\biggr)^{\frac{x-2}{3}\cdot 6+3}\\
&=\lim_{x\rightarrow -\infty}\Biggl[\biggl(1+\frac{1}{\frac{x-2}{3}}\biggr)^{\frac{x-2}{3}}\Biggr]^6\cdot\biggl(1+\frac{1}{\frac{x-2}{3}}\biggr)^3\\
&=e^6.
\end{split}
\]

从而第二题为

\[
\lim_{x\rightarrow -\infty}e^{(\frac{x+1}{x-2})^{2x-1}}=e^{\lim_{x\rightarrow -\infty}(\frac{x+1}{x-2})^{2x-1}}=e^{e^6}.
\]