Cho hàm số $f(x)=x^2-3x+2$, $g(x)=ax+b$ (với a,b là hằng số, $a\ne 0$)

a) ${\lim }_{x\to 0}f(x)=0-3.0+2=2$ đúng

b) ${\lim }_{x\to 2}\frac{f(x)}{x-2}={\lim }_{x\to 2}\frac{x^2-3x+2}{x-2}={\lim }_{x\to 2}\frac{(x-2)(x-1)}{x-2}={\lim }_{x\to 2}(x-1)=1$. Vậy ý b sai.

c) ${\lim }_{x\to +\infty }\frac{\sqrt{f(x)}}{g(x)}={\lim }_{x\to +\infty }\frac{\sqrt{x^2-3x+2}}{ax+b}=\frac{1}{a}$ đúng

d) Để $\underset{x\to 2}{\lim }\frac{f(x)}{ax+b}=2$ thì $ax+b$ có nghiệm bằng $2\iff 2a+b=0\iff b=-2a$; khi đó $\underset{x\to 2}{\lim }\frac{f(x)}{ax+b}=\underset{x\to 2}{\lim }\frac{(x-2)(x-1)}{a(x-2)}=\underset{x\to 2}{\lim }\frac{x-1}{a}=\frac{1}{a}=2\iff a=\frac{1}{2}\Rightarrow b=-1$. Vậy $a+b=-\frac{1}{2}$, ý d sai.
a) ${\lim }_{x\to 0}f(x)=0-3.0+2=2$ đúng
b) ${\lim }_{x\to 2}\frac{f(x)}{x-2}={\lim }_{x\to 2}\frac{x^2-3x+2}{x-2}={\lim }_{x\to 2}\frac{(x-2)(x-1)}{x-2}={\lim }_{x\to 2}(x-1)=1$. Vậy ý b sai.
c) ${\lim }_{x\to +\infty }\frac{\sqrt{f(x)}}{g(x)}={\lim }_{x\to +\infty }\frac{\sqrt{x^2-3x+2}}{ax+b}=\frac{1}{a}$ đúng
d) Để $\underset{x\to 2}{\lim }\frac{f(x)}{ax+b}=2$ thì:
+ $ax+b$ có nghiệm bằng $2\iff 2a+b=0\iff b=-2a$.
+ Khi đó: $\underset{x\to 2}{\lim }\frac{f(x)}{ax+b}=\underset{x\to 2}{\lim }\frac{(x-2)(x-1)}{a(x-2)}=\underset{x\to 2}{\lim }\frac{x-1}{a}=\frac{1}{a}=2\iff a=\frac{1}{2}\Rightarrow b=-1$.
Vậy $a+b=-\frac{1}{2}$, ý d sai.