Cho hàm số $f(x)=x^2-5x+4$, $g(x)=ax+b$ (a,b là hằng số khác 0).

a) ${\lim }_{x\to -1}f(x)={\lim }_{x\to -1}(x^2-5x+4)=10$. Vậy a sai

b) ${\lim }_{x\to 4}\frac{f(x)}{\sqrt{x}-2}={\lim }_{x\to 4}\frac{(x-1)(x-4)\left(\sqrt{x}+2\right)}{x-4}={\lim }_{x\to 4}(x-1)\left(\sqrt{x}+2\right)=12$. Vậy b đúng.

c) ${\lim }_{x\to -\infty }\frac{\sqrt{f(x)}}{g(x)}={\lim }_{x\to -\infty }\frac{\sqrt{x^2-5x+4}}{ax+b}=-\frac{1}{a}$. Vậy c sai

d) Để $\underset{x\to 4}{\lim }\frac{f(x)}{ax+b}=4$ thì $ax+b$ có nghiệm bằng $4\iff 4a+b=0\iff b=-4a$; $\underset{x\to 4}{\lim }\frac{f(x)}{ax+b}=\underset{x\to 4}{\lim }\frac{(x-4)(x-1)}{a(x-4)}=\underset{x\to 4}{\lim }\frac{(x-1)}{a}=\frac{3}{a}=4\iff a=\frac{3}{4}\Rightarrow b=-3\Rightarrow 4a+2b=-3$. Vậy d đúng.
a) ${\lim }_{x\to -1}f(x)={\lim }_{x\to -1}(x^2-5x+4)=10$. Vậy a sai
b) ${\lim }_{x\to 4}\frac{f(x)}{\sqrt{x}-2}={\lim }_{x\to 4}\frac{(x-1)(x-4)\left(\sqrt{x}+2\right)}{x-4}={\lim }_{x\to 4}(x-1)\left(\sqrt{x}+2\right)=12$. Vậy b đúng.
c) ${\lim }_{x\to -\infty }\frac{\sqrt{f(x)}}{g(x)}={\lim }_{x\to -\infty }\frac{\sqrt{x^2-5x+4}}{ax+b}=-\frac{1}{a}$. Vậy c sai
d) Để ${\lim }_{x\to 4}\frac{f(x)}{ax+b}=4$ thì $ax+b$ có nghiệm bằng $4\iff 4a+b=0\iff b=-4a$; ${\lim }_{x\to 4}\frac{f(x)}{ax+b}={\lim }_{x\to 4}\frac{(x-4)(x-1)}{a(x-4)}={\lim }_{x\to 4}\frac{(x-1)}{a}=\frac{3}{a}=4\iff a=\frac{3}{4}\Rightarrow b=-3\Rightarrow 4a+2b=-3$. Vậy d đúng.