Giải thích các bước giải:
Để $\lim_{x\to2}\dfrac{\sqrt{4x-7}+ax+b}{x^2-4x+4}=c$
$\to \lim_{x\to2}\dfrac{(\sqrt{4x-7}-1)+ax+b+1}{(x-2)^2}=c$
$\to \lim_{x\to2}\dfrac{\dfrac{4x-7-1}{\sqrt{4x-7}+1}+a(x-2)+2a+b+1}{(x-2)^2}=c$
$\to \lim_{x\to2}\dfrac{\dfrac{4(x-2)}{\sqrt{4x-7}+1}+a(x-2)+2a+b+1}{(x-2)^2}=c$
$\to 2a+b+1=0 (*)$
$\to \lim_{x\to2}\dfrac{\dfrac{4(x-2)}{\sqrt{4x-7}+1}+a(x-2)}{(x-2)^2}=c$
$\to \lim_{x\to2}\dfrac{\dfrac{4}{\sqrt{4x-7}+1}+a}{x-2}=c$
$\to \lim_{x\to2}\dfrac{4+a\sqrt{4x-7}}{(x-2)(\sqrt{4x-7}+1)}=c$
$\to \lim_{x\to2}\dfrac{4+a+a(\sqrt{4x-7}-1)}{(x-2)(\sqrt{4x-7}+1)}=c$
$\to \lim_{x\to2}\dfrac{4+a+a.\dfrac{4(x-2)}{\sqrt{4x-7}+1}}{(x-2)(\sqrt{4x-7}+1)}=c$
$\to 4+a=0\to a=-4\to b=-1-2a=7$
$\to \lim_{x\to2}\dfrac{-4\dfrac{4}{\sqrt{4x-7}+1}}{\sqrt{4x-7}+1}=c$
$\to \dfrac{-4\dfrac{4}{\sqrt{4.2-7}+1}}{\sqrt{4.2-7}+1}=c$
$\to c=-4$
