\(\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}-\dfrac{3x}{x^2}}+\dfrac{ax}{x}}{\dfrac{bx}{x}-\dfrac{1}{x}}=\dfrac{a-1}{b}=3\)

=> A

1.

\(\lim\dfrac{5\sqrt{3n^2+n}}{2\left(3n+2\right)}=\lim\dfrac{5\sqrt{3+\dfrac{1}{n}}}{2\left(3+\dfrac{2}{n}\right)}=\dfrac{5\sqrt{3}}{6}\Rightarrow a+b=11\)

2.

\(\lim\limits_{x\rightarrow2}\dfrac{x^2+ax+b}{x-2}=6\) khi \(x^2+ax+b=0\) có nghiệm \(x=2\)

\(\Rightarrow4+2a+b=0\Rightarrow b=-2a-4\)

\(\lim\limits_{x\rightarrow2}\dfrac{x^2+ax-2a-4}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)+a\left(x-2\right)}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+a+2\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\left(x+a+2\right)=a+4\Rightarrow a+4=6\Rightarrow a=2\Rightarrow b=-8\)

\(\Rightarrow a+b=-6\)

Đáp án C

a x 2 − 7 x + 12 − b x 2 − 4 x + 3 = a x − 3 x − 4 − b x − 1 x − 3 = a x − 1 − b x − 4 x − 1 x − 3 x − 4

lim x → 3 − x − 1 x − 3 x − 4 = 0

lim x → 3 − a x 2 − 7 x + 12 − b x 2 − 4 x + 3

hữu hạn thì  2 a + b = 0   . Vậy C đúng

Do giới hạn hữu hạn nên \(x^2+mx+n=0\) có nghiệm \(x=1\)

\(\Rightarrow1+m+n=0\Rightarrow n=-m-1\)

\(\lim\limits_{x\rightarrow1}\dfrac{x^2+mx-m-1}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)+m\left(x-1\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1+m\right)}{x-1}=\lim\limits_{x\rightarrow1}\left(x+1+m\right)=m+2\)

\(\Rightarrow m+2=3\Rightarrow m=1\Rightarrow n=-2\)

Đáp án D

lim x → 0 1 − a x + 1 sin b x = lim x → 0 b x sin b x . 1 − a x + 1 b x = lim x → 0 b x sin b x . − a x b x 1 + a x + 1 = lim x → 0 b x sin b x . − a b 1 + a x + 1 = − a 2 b

Chọn C