Question

If $x^2+x^3\le x+f\left(x\right)x^5-x^3$ for values of $x$ near $0,$ then $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}$ is

• A. -1
• B. 0
• C. 1
• D. none of these

Answer 1

Answer: (D)

none of these

$x^2+x^3\le x+f\left(x\right)x^5-x^3$
$x^2+2x^3-x\le f\left(x\right)x^5$
$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}\ge \frac{x^2+2x^3-x}{x^6}$
$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}\ge \frac{x+2x^2-1}{x^5}$
Clearly, the limit does not exist