Đáp án:
$\dfrac1e$
Giải thích các bước giải:
$L = \lim\limits_{x\to \infty}\left(\dfrac{2x +1}{2x+3}\right)^x$
$\to L=\lim\limits_{x\to \infty}e^{\displaystyle{\ln\left(\dfrac{2x +1}{2x+3}\right)^x}}$
$\to L= e^{\displaystyle{\lim\limits_{x\to \infty}\left[x\ln\left(\dfrac{2x +1}{2x+3}\right)\right]}}$
Xét: $\lim\limits_{x\to \infty}\displaystyle{\left[x\ln\left(\dfrac{2x +1}{2x+3}\right)\right]}$
$= \lim\limits_{x\to \infty}\dfrac{\ln\left(\dfrac{2x +1}{2x+3}\right)}{\dfrac1x}$
$=\lim\limits_{x\to \infty}\dfrac{-4x^2}{(2x+1)(2x+3)}$
$=\lim\limits_{x\to \infty}\dfrac{-4}{\left(2+\dfrac1x\right)\left(2 +\dfrac3x\right)}$
$= -1$
Do đó ta được:
$L = e^{-1} =\dfrac1e$