Đáp án: $\lim_{x\to+\infty}\left(\dfrac{x-2}{x+1}\right)^{3x-4}=e^{-9}$
Giải thích các bước giải:
$\lim_{x\to+\infty}\left(\dfrac{x-2}{x+1}\right)^{3x-4}$
$=\lim_{x\to+\infty}\left(\dfrac{x+1-3}{x+1}\right)^{3x-4}$
$=\lim_{x\to+\infty}\left(1+\dfrac{-3}{x+1}\right)^{3\left(x+1\right)-7}$
$=\lim_{x\to+\infty}\left(1+\dfrac{-3}{x+1}\right)^{3\left(x+1\right)}\cdot\left(1+\dfrac{-3}{x+1}\right)^{-7}$
$=\lim_{x\to+\infty}\left(1+\dfrac{-3}{x+1}\right)^{\dfrac{x+1}{-3}\cdot \left(-9\right)}\cdot\left(1+\dfrac{-3}{x+1}\right)^{-7}$
$=\lim_{x\to+\infty}\left(\left(1+\dfrac{-3}{x+1}\right)^{\dfrac{x+1}{-3}}\right)^{-9}\cdot\left(1+\dfrac{-3}{x+1}\right)^{-7}$
$=e^{-9}\cdot\left(1+0\right)^{-7}$
$=e^{-9}$
