question 1 - no calculator.\nlim(x→∞) (x(2x - 3)(8x - 1))/(8x³ - 1)\n(a) -∞\n(b) 1/8\n(c) 1\n(d) 2\n(e) 8

Explanation:

Step1: Expand the numerator

First, expand $x(2x - 3)(8x-1)$. [ \begin{align*} x(2x - 3)(8x-1)&=x(16x^{2}-2x-24x + 3)\ &=x(16x^{2}-26x + 3)\ &=16x^{3}-26x^{2}+3x \end{align*} ]

Step2: Analyze the limit as $x\to\infty$

We want to find $\lim_{x\to\infty}\frac{16x^{3}-26x^{2}+3x}{8x^{3}-1}$. Divide both the numerator and denominator by $x^{3}$: [ \begin{align*} \lim_{x\to\infty}\frac{16x^{3}-26x^{2}+3x}{8x^{3}-1}&=\lim_{x\to\infty}\frac{16-\frac{26}{x}+\frac{3}{x^{2}}}{8-\frac{1}{x^{3}}}\ \end{align*} ] As $x\to\infty$, $\frac{26}{x}\to0$, $\frac{3}{x^{2}}\to0$ and $\frac{1}{x^{3}}\to0$.

Step3: Calculate the limit value

[ \begin{align*} \lim_{x\to\infty}\frac{16-\frac{26}{x}+\frac{3}{x^{2}}}{8-\frac{1}{x^{3}}}&=\frac{16 - 0+0}{8-0}\ &= 2 \end{align*} ]

Answer:

D. 2