Question

Evaluate: $limi{t}_{x\to \infty }\left(\frac{x^3}{3x^2-4}-\frac{x^2}{3x+2}\right)$

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

Answer 1

The correct option is D

$\frac{2}{9}$

Explanation for the correct option:

Evaluate the limit.

$limi{t}_{x\to \infty }\left(\frac{x^3}{3x^2-4}-\frac{x^2}{3x+2}\right)=$

when we substitute limit it becomes the value $\infty$, that does not exist, hence

$$\begin{gathered} limi{t}_{x\to \infty }\left(\frac{x^3}{3x^2-4}-\frac{x^2}{3x+2}\right)=limi{t}_{x\to \infty }\left(\frac{x^3\left(3x+2\right)-x^2\left(3x^2-4\right)}{\left(3x^2-4\right)\left(3x+2\right)}\right) \\ =limi{t}_{x\to \infty }\left(\frac{\cancel{3x^4}+2x^3-\cancel{3x^4}+4x^2}{9x^3-12x+6x^2-8}\right) \\ =limi{t}_{x\to \infty }\left(\frac{x^3\left(2+{\displaystyle \frac{4}{x^3}}\right)}{x^3\left(9-{\displaystyle \frac{12}{x^2}}+{\displaystyle \frac{6}{x}}-{\displaystyle \frac{8}{x^3}}\right)}\right)\mathbf{[}\mathbf{}\mathbf{\text{Taking}}\mathbf{}{\mathbf{x}}^{\mathbf{3}\mathbf{}}\mathbf{}\mathbf{\text{common in numerator and denominator}}\mathbf{}\mathbf{]} \\ =\frac{\left(2+{\displaystyle \frac{4}{{\infty }^3}}\right)}{\left(9-{\displaystyle \frac{12}{{\infty }^2}}+{\displaystyle \frac{6}{\infty }}-{\displaystyle \frac{8}{{\infty }^3}}\right)} \\ =\frac{2}{9}\left[\because \frac{1}{\infty }=0\right] \end{gathered}$$

Hence, option (D) is the correct answer