Question

a,\(^{lim}_{x->2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}\)

b, \(^{lim}_{x->0}\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)

c, \(^{lim}_{x->1}\frac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}\)

d,\(^{lim}_{x->0}\frac{\sqrt{1+2x}.\sqrt[3]{1+4x}-1}{x}\)

e,\(^{lim}_{x->1}\frac{x^4-1}{x^3-2x^2+x}\)

f,\(^{lim}_{x->1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)\)

Answer 1

https://i.imgur.com/v6W1QWU.jpg

Answer 2

ai giup voi

Answer 3

@nguyenvietlam

Answer 4

Liên hợp dài quá làm biếng, đoạn sau ko viết lim bạn tự hiểu nhé:

\(b=\lim\limits_{x\rightarrow0}\frac{2\left(\sqrt{1+x}-1\right)+2-\sqrt[3]{8-x}}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{2x}{\sqrt{1+x}+1}+\frac{x}{4+4\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\frac{2}{\sqrt{1+x}+1}+\frac{1}{4+4\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}=\frac{2}{1+1}+\frac{1}{4+4\sqrt[3]{8}+\sqrt[3]{8^2}}\)

\(c=\lim\limits_{x\rightarrow1}\frac{\sqrt{5-x^3}-2+2-\sqrt[3]{x^2+7}}{\left(x-1\right)\left(x+1\right)}=lim\frac{\frac{-\left(x-1\right)\left(x^2+x+1\right)}{\sqrt{5-x^3}+2}+\frac{-\left(x-1\right)\left(x+1\right)}{4+4\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)}}}{\left(x-1\right)\left(x+1\right)}\)

\(=lim\frac{\frac{-\left(x^2+x+1\right)}{\sqrt{5-x^3}+2}+\frac{-\left(x+1\right)}{4+4\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)}}}{x+1}\) bạn tự thay số và bấm máy

Answer 5

\(d=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}\left(\sqrt[3]{1+4x}-1\right)+\sqrt{1+2x}-1}{x}=lim\frac{\sqrt{1+2x}.\frac{4x}{\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1}+\frac{2x}{\sqrt{1+2x}+1}}{x}\)

\(=lim\frac{4\sqrt{1+2x}}{\sqrt[3]{\left(1+4x\right)^2}+\sqrt[3]{1+4x}+1}+\frac{2}{\sqrt{1+2x}+1}\) bạn thay số và bấm máy

\(e=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{x\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{\left(x+1\right)\left(x^2+1\right)}{x\left(x-1\right)}=\frac{1}{0}=+\infty\)

\(f=\lim\limits_{x\rightarrow1}\left(\frac{1-x^3-3\left(1-x\right)}{\left(1-x\right)\left(1-x^3\right)}\right)=\lim\limits_{x\rightarrow1}\left[\frac{\left(1-x\right)\left(1+x+x^2\right)-3\left(1-x\right)}{\left(1-x\right)^2\left(1+x+x^2\right)}\right]\)

\(=\lim\limits_{x\rightarrow1}\frac{\left(1-x\right)\left(x^2+x-2\right)}{\left(1-x\right)^2\left(1+x+x^2\right)}=\lim\limits_{x\rightarrow1}\frac{-\left(1-x\right)^2\left(x+2\right)}{\left(1-x\right)^2\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{-x-2}{x^2+x+1}\)

Bạn tự thay số