Question

[Trần Phương Thảo]

a. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}\)

b. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}\)

Answer 1

[Nguyễn Việt Lâm]

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

\(=\lim\limits_{x\rightarrow0}3\left(1+x\right)\left(1+2x\right)+\lim\limits_{x\rightarrow0}2\left(1+x\right)+1=3+2+1=6\)

\(b=\lim\limits_{x\rightarrow0}\frac{\left(x^5+5x^4+10x^3+10x^2+5x+1\right)-\left(1+5x\right)}{x^5+x^2}\)

\(=\lim\limits_{x\rightarrow0}\frac{x^2\left(x^3+5x^2+10\right)}{x^2\left(x^3+1\right)}=\lim\limits_{x\rightarrow0}\frac{x^3+5x^2+10}{x^3+1}=10\)

Answer 2

[Trần Phương Thảo]

tại sao a = \(\lim\limits_{x\rightarrow0}\frac{3x\left(1+x\right)\left(1+2x\right)}{x}\)