Question

Evaluate: $\underset{x\to \infty }{lim}\left(\frac{x^3}{2x^2-1}-\frac{x^2}{2x+1}\right)$

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

Answer 1

The correct option is D $\frac{1}{4}$

Given: $\underset{x\to \infty }{lim}\left(\frac{x^3}{2x^2-1}-\frac{x^2}{2x+1}\right)$

First we simplify the expression,

$=\underset{x\to \infty }{lim}\frac{x^3(2x+1)-x^2(2x^2-1)}{(2x^2-1)(2x+1)}$

$\Rightarrow \underset{x\to \infty }{lim}\frac{2x^4+x^3-2x^4+x^2}{(2x^2-1)(2x+1)}$

$=\underset{x\to \infty }{lim}\frac{x^3+x^2}{(2x^2-1)(2x+1)}$

Dividing both the numerator and denominator by $x^3$

$=\underset{x\to \infty }{lim}\frac{1+\frac{1}{x}}{4+\frac{2}{x}-\frac{2}{x^2}-\frac{1}{x^3}}$

We apply the limit in the above expression to get the answer as,

$=\frac{1}{4}$