Đáp án:
$\lim\limits_{x\to - \infty}\left(\sqrt{x^2 + 4x + 1} + x\right)= -2$
Giải thích các bước giải:
\(\begin{array}{l}
\quad \lim\limits_{x\to - \infty}\left(\sqrt{x^2 + 4x + 1} + x\right)\\
= \lim\limits_{x\to - \infty}\dfrac{\left(\sqrt{x^2 + 4x + 1} + x\right)\left(\sqrt{x^2 + 4x + 1} - x\right)}{\sqrt{x^2 + 4x + 1} - x}\\
= \lim\limits_{x\to - \infty}\dfrac{4x+1}{\sqrt{x^2 + 4x + 1} - x}\\
= \lim\limits_{x\to - \infty}\dfrac{4 + \dfrac1x}{-\sqrt{1+ \dfrac4x + \dfrac{1}{x^2}} - 1}\\
= \dfrac{4+0}{-\sqrt{1+0+0}-1}\\
= -2
\end{array}\)
