Bài 1: Giới hạn của dãy số

NN

Tìm các giới hạn sau :

A=\(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{x+1}-1}{\sqrt[4]{2x+1}-1}\)

B=\(\lim\limits_{x\rightarrow7}\frac{\sqrt[3]{4x-1}\sqrt{x-2}}{\sqrt[4]{2x+2}-2}\)

C=\(\lim\limits_{x\rightarrow0}\frac{\sqrt{\left(2x+1\right)\left(3x+1\right)\left(4x+1\right)}-1}{x}\)

D=\(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}\)

E=\(\lim\limits_{x\rightarrow0}\frac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{x}\)

Giup mình vớiii

NL
2 tháng 4 2020 lúc 10:15

\(A=\lim\limits_{x\rightarrow0}\frac{\left(x+1\right)^{\frac{1}{3}}-1}{\left(2x+1\right)^{\frac{1}{4}}-1}=\lim\limits_{x\rightarrow0}\frac{\frac{1}{3}\left(x+1\right)^{-\frac{2}{3}}}{\frac{1}{2}\left(2x+1\right)^{-\frac{3}{4}}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}\)

\(B=\lim\limits_{x\rightarrow7}\frac{\sqrt[3]{4x-1}\sqrt{x-2}}{\sqrt[4]{2x+2}-2}=\frac{3\sqrt{5}}{0}=+\infty\)

\(C=\lim\limits_{x\rightarrow0}\frac{\sqrt{\left(3x+1\right)\left(4x+1\right)}\left(\sqrt{2x+1}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{4x+1}\left(\sqrt{3x+1}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{4x+1}-1}{x}\)

Xét \(\lim\limits_{x\rightarrow0}\frac{\sqrt{ax+1}-1}{x}=\lim\limits_{x\rightarrow0}\frac{\left(ax+1\right)^{\frac{1}{2}}-1}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{a}{2}\left(ax+1\right)^{-\frac{1}{2}}}{1}=\frac{a}{2}\)

\(\Rightarrow C=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}=\frac{9}{2}\)

\(D=\lim\limits_{x\rightarrow0}\frac{\left(1+4x\right)^{\frac{1}{2}}-\left(1+6x\right)^{\frac{1}{3}}}{x^2}=\lim\limits_{x\rightarrow0}\frac{2\left(1+4x\right)^{-\frac{1}{2}}-2\left(1+6x\right)^{-\frac{2}{3}}}{2x}\)

\(D=\lim\limits_{x\rightarrow0}\frac{-2\left(1+4x\right)^{-\frac{3}{2}}+4\left(1+6x\right)^{-\frac{5}{3}}}{1}=-2+4=2\)

\(E=\lim\limits_{x\rightarrow0}\frac{\left(1+ax\right)^{\frac{1}{n}}-\left(1+bx\right)^{\frac{1}{n}}}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{a}{n}\left(1+ax\right)^{\frac{1-n}{n}}-\frac{b}{n}\left(1+bx\right)^{\frac{1-n}{n}}}{1}=\frac{a-b}{n}\)

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NL
2 tháng 4 2020 lúc 15:39

\(B=\lim\limits_{x\rightarrow7}\frac{\sqrt[3]{4x-1}-\sqrt{x+2}}{\sqrt[4]{2x+2}-2}=\lim\limits_{x\rightarrow7}\frac{\left(4x-1\right)^{\frac{1}{3}}-\left(x+2\right)^{\frac{1}{2}}}{\left(2x+2\right)^{\frac{1}{4}}-2}\)

\(B=\lim\limits_{x\rightarrow7}\frac{\frac{4}{3}\left(4x-1\right)^{-\frac{2}{3}}-\frac{1}{2}\left(x+2\right)^{-\frac{1}{2}}}{\frac{1}{2}\left(2x+2\right)^{-\frac{3}{4}}}=\lim\limits_{x\rightarrow7}\frac{\frac{4}{3\sqrt[3]{\left(4x-1\right)^2}}-\frac{1}{2\sqrt{x+2}}}{\frac{1}{2}\sqrt[4]{\left(2x+2\right)^3}}\)

\(=\frac{\frac{4}{3\sqrt[3]{27^2}}-\frac{1}{2\sqrt{9}}}{\frac{1}{2}\sqrt[4]{16^3}}=-\frac{1}{216}\)

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