Đáp án:
$L = \lim\limits_{x\to \pm \infty}\left(\dfrac{2x+5}{2x-1}\right)^{\displaystyle{\dfrac{x^2+1}{2x+5}}} = \sqrt{e^3}$
Giải thích các bước giải:
$\begin{array}{l}\quad L = \lim\limits_{x\to \pm \infty}\left(\dfrac{2x+5}{2x-1}\right)^{\displaystyle{\dfrac{x^2+1}{2x+5}}}\\ \to L = e^{\displaystyle{\lim\limits_{x\to \pm \infty}\ln\left[\left(\dfrac{2x+5}{2x-1}\right)^{\displaystyle{\dfrac{x^2+1}{2x+5}}}\right]}}\\ \to L = e^{\displaystyle{\lim\limits_{x\to \pm \infty}\left[\dfrac{x^2+1}{2x+5}\ln\left(\dfrac{2x+5}{2x-1}\right)\right]}}\\ Xét\,\,L' = \lim\limits_{x\to \pm \infty}\left[\dfrac{x^2+1}{2x+5}\ln\left(\dfrac{2x+5}{2x-1}\right)\right]\\ \to L' = \lim\limits_{x\to \pm \infty}\left[\dfrac{x^2+1}{2x^2+5x}\cdot x\ln\left(\dfrac{2x+5}{2x-1}\right)\right]\\ \to L' = \lim\limits_{x\to \pm \infty}\dfrac{x^2+1}{2x^2+5x}\cdot\lim\limits_{x\to \pm \infty}\left[x\ln\left(\dfrac{2x+5}{2x-1}\right)\right]\\ \to L' = \lim\limits_{x\to \pm \infty}\dfrac{1+\dfrac{1}{x^2}}{2+\dfrac5x}\cdot\lim\limits_{x\to \pm \infty}\left[\dfrac{\ln\left(\dfrac{2x+5}{2x-1}\right)}{\dfrac1x}\right]\\ \to L' = \dfrac{1+0}{2+0}\cdot\lim\limits_{x\to \pm \infty}\dfrac{-\dfrac{12}{(2x-1)^2}:\dfrac{2x+5}{2x-1}}{-\dfrac{1}{x^2}}\\ \to L' = \dfrac12\cdot\lim\limits_{x\to \pm \infty}\dfrac{12x^2}{(2x-1)(2x+5)}\\ \to L' = 6\cdot\lim\limits_{x\to \pm \infty}\dfrac{x^2}{(2x-1)(2x+5)}\\ \to L' = 6\cdot\lim\limits_{x\to \pm \infty}\dfrac{x}{4(x+1)}\\ \to L' = \dfrac32\cdot\lim\limits_{x\to \pm \infty}\dfrac{x}{x+1}\\ \to L' =\dfrac32\cdot\lim\limits_{x\to \pm \infty}\dfrac{1}{1 + \dfrac1x}\\ \to L' = \dfrac32\cdot\dfrac{1}{1+0} = \dfrac32\\ \to L = e^{\displaystyle{L'}}\\ \to L = e^{\displaystyle{\dfrac32}} = \sqrt{e^3} \end{array}$
