Câu 11:
$\sqrt{x-3}>0$
$\lim\limits_{x\to 3}\Big( \dfrac{\sqrt{x-3}-(x-3)}{x-3}\Big)$
$=\lim\limits_{x\to 3}\Big( \dfrac{\sqrt{x-3}}{x-3}-1\Big)$
$=\lim\limits_{x\to 3}\Big( \dfrac{1}{\sqrt{x-3}}-1)$
$=+\infty$
$\to B$
Câu 12:
$\lim\limits_{x\to 0}\dfrac{\sqrt{x^2+1}-1}{4-\sqrt{x^2+16}}$
$=\lim\limits_{x\to 0}\dfrac{x^2}{(4-\sqrt{x^2+16})(\sqrt{x^2+1}+1)}$
$=\lim\limits_{x\to 0}\dfrac{x^2(4+\sqrt{x^2+16})}{(16-x^2-16)(\sqrt{x^2+1}+1)}$
$=\lim\limits_{x\to 0}\dfrac{4+\sqrt{x^2+16}}{-\sqrt{x^2+1}-1}$
$=\dfrac{4+4}{-1-1}$
$=-4$
Hoành độ đỉnh parabol $y=ax^2+bx+c$ là $\dfrac{-b}{2a}$
Thử từng đáp án, thấy $\dfrac{-16}{2.2}=-4$
$\to A$
