Đáp án:
\(\begin{array}{l}
a)\quad \lim\limits_{x\to 1}\dfrac{x^3 - 1}{x^2 - 1}= \dfrac32\\
b)\quad \lim\limits_{x\to\infty}\dfrac{2x + 1}{3x-1}= \dfrac23\\
c)\quad \lim\limits_{x\to 0^+}x\ln x= 0\\
d)\quad \lim\limits_{x\to \infty}\left(\sqrt{x^2 + 3x + 1} - x\right)= \dfrac32
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\quad \lim\limits_{x\to 1}\dfrac{x^3 - 1}{x^2 - 1}\\
= \lim\limits_{x\to 1}\dfrac{x^2 + x + 1}{x + 1}\\
= \dfrac{1+1+1}{1+1}\\
= \dfrac32\\
b)\quad \lim\limits_{x\to\infty}\dfrac{2x + 1}{3x-1}\\
= \lim\limits_{x\to\infty}\dfrac{2 + \dfrac1x}{3 - \dfrac1x}\\
= \dfrac{2 +0}{3 - 0}\\
= \dfrac23\\
c)\quad \lim\limits_{x\to 0^+}x\ln x\\
= \lim\limits_{x\to 0^+}\dfrac{\ln x}{\dfrac1x}\\
= \lim\limits_{x\to 0^+}\left(\dfrac{\dfrac1x}{-\dfrac{1}{x^2}}\right)\qquad \text{(l'Hôpital)}\\
= \lim\limits_{x\to 0^+}(-x)\\
= 0\\
d)\quad \lim\limits_{x\to \infty}\left(\sqrt{x^2 + 3x + 1} - x\right)\\
= \lim\limits_{x\to \infty}\dfrac{\left(\sqrt{x^2 + 3x + 1} - x\right)\left(\sqrt{x^2 + 3x + 1} +x\right)}{\sqrt{x^2 + 3x + 1} + x}\\
= \lim\limits_{x\to \infty}\dfrac{3x+1}{\sqrt{x^2 + 3x + 1} + x}\\
= \lim\limits_{x\to \infty}\dfrac{3 + \dfrac1x}{\sqrt{1 + \dfrac{3}{x} + \dfrac{1}{x^2}} + 1}\\
= \dfrac{3+0}{\sqrt{1 + 0+0} + 1}\\
= \dfrac32
\end{array}\)
