Đáp án: $-\dfrac{9}{14}$
Giải thích các bước giải:
$\lim_{x\to1}\dfrac{\sqrt{5-x^2}-\sqrt[3]{x+7}}{x-1}$
$=\lim_{x\to1}\dfrac{(\sqrt{5-x^2}-2)-(\sqrt[3]{x+7}-2)}{x-1}$
$=\lim_{x\to1}\dfrac{\dfrac{5-x^2-2^2}{\sqrt{5-x^2}+2}-\dfrac{x+7-2^3}{\sqrt[3]{x+7}^2+\sqrt[3]{x+7}+1}}{x-1}$
$=\lim_{x\to1}\dfrac{\dfrac{(1-x)(1+x)}{\sqrt{5-x^2}+2}-\dfrac{x-1}{\sqrt[3]{x+7}^2+\sqrt[3]{x+7}+1}}{x-1}$
$=\lim_{x\to1}-\dfrac{x+1}{\sqrt{5-x^2}+2}-\dfrac{1}{\sqrt[3]{x+7}^2+\sqrt[3]{x+7}+1}$
$=-\dfrac{1+1}{\sqrt{5-1^2}+2}-\dfrac{1}{\left(\sqrt[3]{1+7}\right)^2+\sqrt[3]{1+7}+1}$
$=-\dfrac{9}{14}$
