Question

${\lim }_{x\to \infty }\frac{3x^3+2x^2-7x+9}{4x^3+9x-2}$ is equal to

• A. $\frac{2}{9}$
• B. $\frac{1}{2}$
• C. $\frac{-9}{2}$
• D. $\frac{3}{4}$
• E. $\frac{9}{2}$

Answer 1

Answer: (D)

$\underset{x\to \infty }{\lim }\frac{3x^3+2x^2-7x+9}{4x^3+9x-2}$
$=\underset{x\to \infty }{\lim }\frac{x^3\left[3+\frac{2}{x}-\frac{7}{x^2}+\frac{9}{x^3}\right]}{x^3\left[4+\frac{9}{x^2}-\frac{2}{x^3}\right]}$
On putting $x\to \infty$, we get
$=\frac{[3+0-0+0]}{[4+0-0]}=\frac{3}{4}$