Question

The value of $\underset{x\to \infty }{lim}\frac{2x^3-4x+7}{3x^3+5x^2-4}$ is

• A. $\frac{2}{3}$
• B. $-\frac{7}{4}$
• C. $-\frac{4}{5}$
• D. $\infty$

Answer 1

Answer: (A)

$\frac{2}{3}$

$\underset{x\to \infty }{lim}\frac{2x^3-4x+7}{3x^3+5x^2-4}$

$=\underset{x\to \infty }{lim}\frac{x^3}{x^3}\times \frac{2-\frac{4}{x^2}+\frac{7}{x^3}}{(3+\frac{5}{x}-\frac{4}{x^3})}$ $as(x\to \infty ;\frac{1}{x}\to 0)$

$=\frac{2-0+0}{3+0-0}=\frac{2}{3}$