Đáp án:
\(\lim\limits_{x\to \infty}\left(\dfrac{3x^2 - x +1}{2x^2 + x +1}\right)^{\dfrac{x^2}{x-1}}= \infty\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \lim\limits_{x\to \infty}\left(\dfrac{3x^2 - x +1}{2x^2 + x +1}\right)^{\dfrac{x^2}{x-1}}\\
= \lim\limits_{x\to \infty}e^{\displaystyle{\ln\left(\dfrac{3x^2 - x +1}{2x^2 + x +1}\right)^{\dfrac{x^2}{x-1}}}}\\
= e^{\displaystyle{\lim\limits_{x\to \infty}}\ln\left(\dfrac{3x^2 - x +1}{2x^2 + x +1}\right)^{\dfrac{x^2}{x-1}}}\\
= e^{\displaystyle{\lim\limits_{x\to \infty}}\left[\dfrac{x^2}{x-1}\ln\left(\dfrac{3x^2 - x +1}{2x^2 + x +1}\right)\right]}\\
= e^{\displaystyle{\lim\limits_{x\to \infty}}\dfrac{x^2}{x-1}\cdot \lim\limits_{x\to \infty}\ln\left(\dfrac{3x^2 - x +1}{2x^2 + x +1}\right)}\\
= e^{\displaystyle{\infty\cdot \dfrac32}}\\
= \infty
\end{array}\)
