Question

Tìm giới hạn

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\left(x^2-1\right)\left(1-2x\right)^5}{x^7+x+3}\)

Answer 1

Lời giải:

Ta có: \(\lim _{x\to -\infty}\frac{(x^2-1)(1-2x)^5}{x^7+x+3}=\lim_{x\to -\infty}\frac{\frac{(x^2-1)(1-2x)^5}{x^7}}{\frac{x^7+x+3}{x^7}}\)

\(=\lim_{x\to -\infty}\frac{\left ( \frac{x^2-1}{x^2} \right )\left ( \frac{1-2x}{x} \right )^5}{1+\frac{1}{x^6}+\frac{3}{x^7}}=\lim_{x\to -\infty}\frac{\left ( 1-\frac{1}{x^2} \right )\left ( \frac{1}{x}-2 \right )^5}{1+\frac{1}{x^6}+\frac{3}{x^7}}\)

\(=\frac{1(-2)^5}{1}=-32\)

(Nhớ rằng \(\lim_{x\to \infty}\frac{1}{x}=0\) là được )