How to find $\lim_{x\rightarrow\infty}\bigl(\frac{x+1}{x-2}\bigr)^{2x-1}$

Question

$$\lim_{x\rightarrow\infty}\left(\frac{x+1}{x-2}\right)^{2x-1}$$

What are the steps to solve it? Probably the division should be multiplied by some expression.

Answer 1

$$\left(\frac{1+x}{-2+x}\right)^{-1+2x}=\left[\left(1+\frac3{x-2}\right)^{x-2}\right]^{2}\left(1+\frac3{x-2}\right)^3$$

And now use the basic lemma

$$\lim_{x\to\infty}\left(1+\frac a{f(x)}\right)^{f(x)}=e^a$$

where $\;a\in\Bbb R\;$ and $\;f\;$ is a funcion s.t.

$$\lim_{x\to\infty}f(x)=\infty$$

Answer 2

As $\displaystyle \frac{x+1}{x-2}=1+\frac3{x-2},$

$$\lim_{x\to\infty}\left(\frac{1+x}{-2+x}\right)^{-1+2x}$$

$$=\left(\lim_{x\to\infty}\left(1+\frac3{-2+x}\right)^{\frac{x-2}3}\right)^{\lim_{x\to\infty}\frac{3(2x-1)}{x-2}}$$

$$=e^6$$ as $\lim_{x\to\infty}\frac{3(2x-1)}{x-2}=3\lim_{x\to\infty}\frac{2-\frac1x}{1-\frac2x}=6$ and $\lim_{y\to\infty}\left(1+\frac1y\right)^y=e$