Đáp án: $\dfrac34$
Giải thích các bước giải:
Vì $x\to 3^+\to |3-x|=|x-3|=x-3$
Ta có :
$\lim_{x\to 3^+}\dfrac{|3-x|\sqrt{x^2+7}-4(x-3)}{(x-3)^2}$
$=\lim_{x\to 3^+}\dfrac{(x-3)\sqrt{x^2+7}-4(x-3)}{(x-3)^2}$
$=\lim_{x\to 3^+}\dfrac{\sqrt{x^2+7}-4}{x-3}$
$=\lim_{x\to 3^+}\dfrac{(\sqrt{x^2+7}-4)(\sqrt{x^2+7}+4)}{(x-3)(\sqrt{x^2+7}+4)}$
$=\lim_{x\to 3^+}\dfrac{x^2+7-4^2}{(x-3)(\sqrt{x^2+7}+4)}$
$=\lim_{x\to 3^+}\dfrac{x^2-9}{(x-3)(\sqrt{x^2+7}+4)}$
$=\lim_{x\to 3^+}\dfrac{(x-3)(x+3)}{(x-3)(\sqrt{x^2+7}+4)}$
$=\lim_{x\to 3^+}\dfrac{x+3}{\sqrt{x^2+7}+4}$
$=\dfrac{3+3}{\sqrt{3^2+7}+4}$
$=\dfrac34$
