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)

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