question 1 - no calculator.\nlim(x→∞) (x(2x -...
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
Answer
# 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