Question

$\underset{x\to \infty }{\lim }$ $\left(\frac{x^3}{3x^2-4}-\frac{x^2}{3x+2}\right)$ is equal to

• A. $\frac{-1}{4}$
• B. $\frac{-1}{2}$
• C. $0$
• D. $\frac{2}{9}$

Answer 1

Answer: (D)

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

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

$=\underset{x\to \infty }{\lim }$$\left(\frac{3x^4+2x^3-3x^4+4x^2}{9x^3+6x^2-12x-8}\right)$

$=\underset{x\to \infty }{\lim }$$\left(\frac{2x^3+4x^2}{9x^3+6x^2-12x-8}\right)$

$=\underset{x\to \infty }{\lim }$$\left(\frac{2+\frac{4}{x}}{9+\frac{6}{x}-\frac{12}{x^2}-\frac{8}{x^3}}\right)$

$=\frac{2}{9}\left[\because \underset{x\to \infty }{\lim }\frac{a}{x}=0\right]$