Đáp án:
Giải thích các bước giải:
b. $\lim_{x \to -2}\frac{4-x^2}{x+2} $ = $\lim_{x \to -2}\frac{(2-x)(2+x)}{x+2} $ = $\lim_{x \to -2} 2-x $ = 4
f. $\lim_{x \to 2}\frac{2-x}{\sqrt[]{x+7}-3} $ = $\lim_{x \to 2}\frac{(2-x)(\sqrt[]{x+7}+3)}{x+7-9} $= $\lim_{x \to 2}\frac{(2-x)(\sqrt[]{x+7}+3)}{x-2} $ = $\lim_{x \to 2}-(\sqrt[]{x+7}+3) $ = -6
j. $\lim_{x \to -2}\frac{\sqrt[]{x^2+5}-3}{x+2} $ = $\lim_{x \to -2}\frac{x^2+5-9}{(x+2)(\sqrt[]{x^2+5}+3)} $ = $\lim_{x \to -2}\frac{x^2-4}{(x+2)(\sqrt[]{x^2+5}+3)} $ = $\lim_{x \to -2}\frac{x-2}{\sqrt[]{x^2+5}+3} $ = $\frac{-2}{3}$
