$B=\lim_{x \to 1} $ $\frac{\sqrt[]{1+3x}-\sqrt[3]{1+7x}}{x-1}$
$B=\lim_{x \to 1} $ $\frac{(\sqrt[]{1+3x}-2)+(2-\sqrt[3]{1+7x})}{x-1}$
$B=\lim_{x \to 1} $ $\frac{\sqrt[]{1+3x}-2}{x-1}+$ $\lim_{x \to 1} $ $\frac{2-\sqrt[3]{1+7x}}{x-1}$
Đặt $B_1=\lim_{x \to 1} $ $\frac{\sqrt[]{1+3x}-2}{x-1},$ $B_2=\lim_{x \to 1} $ $\frac{2-\sqrt[3]{1+7x}}{x-1}$
$B_1=\lim_{x \to 1} $ $\frac{\sqrt[]{1+3x}-2}{x-1}$
$=\lim_{x \to 1} $$\frac{1+3x-4}{(x-1)(\sqrt[]{1+3x}+2)}$
$=\lim_{x \to 1} $$\frac{3(x-1)}{(x-1)(\sqrt[]{1+3x}+2)}$
$=\lim_{x \to 1} $$\frac{3}{\sqrt[]{1+3x}+2}=$ $\frac{3}{4}$
$B_2=\lim_{x \to 1} $ $\frac{2-\sqrt[3]{1+7x}}{x-1}$
$=\lim_{x \to 1} $ $\frac{8-1-7x}{(x-1)[4+2\sqrt[3]{1+7x}+(\sqrt[3]{1+7x})^2]}$
$=\lim_{x \to 1} $ $\frac{7(1-x)}{(x-1)[4+2\sqrt[3]{1+7x}+(\sqrt[3]{1+7x})^2]}$
$=\lim_{x \to 1} $ $\frac{-7}{4+2\sqrt[3]{1+7x}+(\sqrt[3]{1+7x})^2}=$$\frac{-7}{12}$
Vậy $B=B_1+B_2=$$\frac{3}{4}+$ $(\frac{-7}{12})=$ $\frac{1}{6}$
