Đáp án:
\(\begin{array}{l}
a)\quad \lim\limits_{x\to +\infty}\dfrac{x + \sqrt{x^2 +1}}{x + \sqrt[3]{x^3 + 1}}= 1\\
b)\quad \lim\limits_{x\to -\infty}\dfrac{\sqrt{x^2 + 1}}{x-1}= -1\\
c)\quad \lim\limits_{x\to -\infty}(x^3 + 1)=- \infty
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\quad \lim\limits_{x\to +\infty}\dfrac{x + \sqrt{x^2 +1}}{x + \sqrt[3]{x^3 + 1}}\\
= \lim\limits_{x\to +\infty}\dfrac{1 + \sqrt{1 + \dfrac{1}{x^2}}}{1 + \sqrt[3]{1 + \dfrac{1}{x^3}}}\\
= \dfrac{1 + \sqrt{1 + 0}}{1 + \sqrt[3]{1 + 0}}\\
= 1\\
b)\quad \lim\limits_{x\to -\infty}\dfrac{\sqrt{x^2 + 1}}{x-1}\\
=\lim\limits_{x\to -\infty}\dfrac{-\sqrt{1 + \dfrac{1}{x^2}}}{1 - \dfrac{1}{x}}\\
= \dfrac{-\sqrt{1 + 0}}{1 - 0}\\
= -1\\
c)\quad \lim\limits_{x\to -\infty}(x^3 + 1)\\
= -\infty + 1\\
= - \infty
\end{array}\)
