Question

Answer 1

Lời giải:

\(\lim\limits_{x\to 0}\frac{(3x+1)^3-(4x-1)^4}{x}=\lim\limits_{x\to 0}\frac{(3x+1)^3-1-[(4x-1)^4-1]}{x}=\lim\limits_{x\to 0}\frac{3x[(3x+1)^2+(3x+1)+1]-4x(4x-2)[(4x-1)^2+1]}{x}\)

\(=\lim\limits_{x\to 0}[3(3x+1)^2+3(3x+1)+3-4(4x-2)((4x-1)^2+1)]=25\)