Question

$\underset{x\to \infty }{\lim }\frac{\left(3x-1\right)\left(4x-2\right)}{\left(x+8\right)\left(x-1\right)}$

Answer 1

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{\left(3x-1\right)\left(4x-2\right)}{\left(x+8\right)\left(x-1\right)} \\ \mathrm{Dividing}\mathrm{the}\mathrm{numerator}\mathrm{and}\mathrm{the}\mathrm{denominator}\mathrm{by}{x}^2: \\ \underset{x\to \infty }{\lim }\frac{\left({\displaystyle \frac{3x-1}{x}}\right)\left({\displaystyle \frac{4x-2}{x}}\right)}{\left({\displaystyle \frac{x+8}{x}}\right)\left({\displaystyle \frac{x-1}{x}}\right)} \\ =\underset{x\to \infty }{\lim }\left[\frac{\left(3-{\displaystyle \frac{1}{x}}\right)\left(4-{\displaystyle \frac{2}{x}}\right)}{\left(1+{\displaystyle \frac{8}{x}}\right)\left(1-{\displaystyle \frac{1}{x}}\right)}\right] \\ \mathrm{When}x\to \infty ,\mathrm{then}\frac{1}{x}\to 0. \\ \frac{3\times 4}{1} \\ =12 \end{gathered}$$