Đáp án:
a,$3\sqrt{2}$
b,$\frac{7}{96}$
Giải thích các bước giải:
a,$\lim\limits_{x\rightarrow -\infty} (3x+1).\sqrt{\frac{2x+1}{3+x^3}}$
=$\lim\limits_{x\rightarrow -\infty }{x.(3+\frac{1}{x}).\sqrt{\frac{2+\frac{1}{x}}{\frac{3}{x}+x^2}}}$
=$\lim\limits_{x\rightarrow -\infty }{-x.(3+\frac{1}{x}).\frac{1}{-x}.\sqrt{\frac{2+\frac{1}{x}}{\frac{3}{x^3}+1}}}$
=$\lim\limits_{x\rightarrow -\infty}{(3+\frac{1}{x} ).\sqrt{\frac{2+\frac{1}{x}}{\frac{3}{x^3}+1}}}$
=$3\sqrt{2}$
b,$\lim\limits_{x\rightarrow 1}{\frac{\sqrt[3]{x+7}-\sqrt{3x+1}}{3x^2+x-4}}$
=$\lim\limits_{x\rightarrow 1}{\frac{\sqrt[3]{x+7}-2+2-\sqrt{3x+1}}{(3x+4).(x-1)}}$
=$\lim\limits_{x\rightarrow 1}{\frac{1}{(3x+4).(x-1)}.[\frac{x-1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}+\frac{3(x-1)}{4+\sqrt{3x+1)}}]}$
=$\lim\limits_{x\rightarrow 1}{\frac{1}{3x+4}.[\frac{1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}+\frac{3}{4+\sqrt{3x+1}}]}$
=$\frac{1}{8}.(\frac{1}{4+4+4}+\frac{3}{4+2})$
=$\frac{7}{96}$
