Lời giải:
$\lim\limits_{x \to -\infty} (\sqrt[]{4x^2-8x+10}+2x)$
$=\lim\limits_{x \to -\infty} \dfrac{(\sqrt[]{4x^2-8x+10}+2x)(\sqrt[]{4x^2-8x+10}-2x)}{\sqrt[]{4x^2-8x+10}-2x}$
$=\lim\limits_{x \to -\infty} \dfrac{(4x^2-8x+10)-(2x)^2}{\sqrt[]{4x^2-8x+10}-2x}$
$=\lim\limits_{x \to -\infty} \dfrac{-8x+10}{\sqrt[]{4x^2-8x+10}-2x}$
$=\lim\limits_{x \to -\infty} \dfrac{-8+\dfrac{10}{x}}{-\sqrt[]{4-\dfrac{8}{x}+\dfrac{10}{x^2}}-2}$
$=\dfrac{-8}{-4}=2$
Giải thích:
Do $\sqrt{A^2}=|A|>0$ $(\forall A)$
$\to$ khi $x\to-\infty$ thì $x=-\sqrt{x^2}$
