Kết quả nào dưới đây là sai ?
\(\lim\limits_{x\rightarrow1}\dfrac{x^3-1}{x^3-x^2+x-1}=\dfrac{3}{2}\).\(\lim\limits_{x\rightarrow-3}\dfrac{x^4-6x^2-27}{x^3+3x^2+x+3}=-\dfrac{36}{5}\).\(\lim\limits_{x\rightarrow0}\dfrac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}=5\).\(\lim\limits_{x\rightarrow-1}=\dfrac{x^5+1}{x^3+1}=\dfrac{5}{3}\).Hướng dẫn giải:Cách 1 (biến đổi hàm số cần tính giới hạn):
\(\frac{x^3-1}{x^3-x^2+x-1}=\frac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+1\right)}=\frac{x^2+x+1}{x^2+1}\)\(\Rightarrow\lim\limits_{x\rightarrow1}\frac{x^3-1}{x^3-x^2+x-1}=\lim\limits_{x\rightarrow1}\frac{x^2+x+1}{x^2+1}=\frac{3}{2}\)
\(\frac{x^4-6x^2-27}{x^3+3x^2+x+3}=\frac{\left(x^2-9\right)\left(x^2+3\right)}{\left(x+3\right)\left(x^2+1\right)}=\frac{\left(x-3\right)\left(x^2+3\right)}{x^2+1}\)\(\Rightarrow\lim\limits_{x\rightarrow-3}\frac{x^4-6x^2-27}{x^3+3x^2+x+3}=\lim\limits_{x\rightarrow-3}\frac{\left(x-3\right)\left(x^2+3\right)}{x^2+1}=-\frac{36}{5}\)
\(\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}=\frac{6x^3+6x^2+6x}{x}=6x^2+6x+6\) \(\Rightarrow\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}=\lim\limits_{x\rightarrow0}\left(6x^2+6x+6\right)=6\ne5\)
Vậy khẳng định sai là \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}=5\)
