Giải thích các bước giải:
Ta có :
$\lim_{x\to2^+}\dfrac{2-x}{\sqrt{x+7}-3}=\lim_{x\to2^+}\dfrac{2-x}{\dfrac{x+7-3^2}{\sqrt{x+7}+3}}$
$\to\lim_{x\to2^+}\dfrac{2-x}{\sqrt{x+7}-3}=\lim_{x\to2^+}\dfrac{2-x}{\dfrac{x-2}{\sqrt{x+7}+3}}$
$\to\lim_{x\to2^+}\dfrac{2-x}{\sqrt{x+7}-3}=\lim_{x\to2^+}-(\sqrt{x+7}+3)$
$\to\lim_{x\to2^+}\dfrac{2-x}{\sqrt{x+7}-3}=-(\sqrt{2+7}+3)=-6$
$\lim_{x\to2^-}\dfrac{x^2-10x+16}{x-2}=\lim_{x^-\to2}\dfrac{(x-2)(x-8)}{x-2}=\lim_{x\to2^-}x-8=-6$
$\to \lim_{x\to2^+}\dfrac{2-x}{\sqrt{x+7}-3}=f(2)=\lim_{x\to2^-}\dfrac{x^2-10x+16}{x-2}$
$\to$Hàm số liên tục
