Question

$\underset{x\to \infty }{\lim }\frac{3x^3-4x^2+6x-1}{2x^3+x^2-5x+7}$

Answer 1

$$\begin{gathered} \underset{x\to \infty }{\lim }\left[\frac{3x^3-4x^2+6x-1}{2x^3+x^2-5x+7}\right] \\ \mathrm{Dividing}\mathrm{the}\mathrm{numerator}\mathrm{and}\mathrm{the}\mathrm{denominator}\mathrm{by}{x}^3: \\ \underset{x\to \infty }{\lim }\left[\frac{3-{\displaystyle \frac{4}{x}}+{\displaystyle \frac{6}{x^2}}-{\displaystyle \frac{1}{x^3}}}{2+{\displaystyle \frac{1}{x}}-{\displaystyle \frac{5}{x^2}}+{\displaystyle \frac{7}{x^3}}}\right] \\ \mathrm{When}x\to \infty ,\mathrm{then}\frac{1}{x},\frac{1}{x^2},\frac{1}{x^3}\to 0. \\ \Rightarrow \frac{3}{2} \end{gathered}$$