Câu 1:
$\lim\limits_{x\to 2} \dfrac{4-x^2}{2-x}=\lim\limits_{x\to 2}\dfrac{(2-x)(2+x)}{2-x}=\lim\limits_{x\to 2} (x+2)=4$
Câu 2:
$\lim\limits_{x\to 6}\dfrac{\sqrt{x+3}-3}{x-6}=\lim\limits_{x\to 6}\dfrac{\big(\sqrt{x+3}-3\big)\big(\sqrt{x+3}+3\big)}{(x-6)\big(\sqrt{x+3}+3\big)}=\lim\limits_{x\to 6}\dfrac{x-6}{(x-6)\big(\sqrt{x+3}+3\big)}=\lim\limits_{x\to 6}\dfrac1{\sqrt{x+3}+3}=\dfrac16$
Câu 3:
$\lim\big(-x^3+2x^2+x-1\big)=\lim x^3\left(-1+\dfrac2x+\dfrac1{x^2}-\dfrac1{x^3}\right)\\\text{Ta có:}\\\lim x^3=+\infty\\\lim \left(-1+\dfrac2x+\dfrac1{x^2}-\dfrac1{x^3}\right)=-1\\\Rightarrow \lim\big(-x^3+2x^2+x-1\big)=-\infty$
