Question

In Problems 33– 44, find the one-sided limit.

$\underset{x\to 0^{+}}{\lim }\frac{x^3-x^2}{x^4+x^2}$

Short answer

the limit value

$\underset{x\to 0^{+}}{\lim }\frac{x^3-x^2}{x^4+x^2}=-1$

Step-by-step solution

step 1.introduction

given limit

$\underset{x\to 0^{+}}{\lim }\frac{x^3-x^2}{x^4+x^2}$

step 2.equation simplification

consider the equation

$$\begin{gathered} \frac{x^3-x^2}{x^4+x^2}=\frac{x^2(x-1)}{x^2(x^2+1)} \\ =\frac{x-1}{x^2+1} \end{gathered}$$

step 3.the solution

the limit values

$$\begin{gathered} \frac{x^3-x^2}{x^4+x^2}=\underset{x\to 0^{+}}{\lim }\frac{x-1}{x^2+1} \\ =\frac{0-1}{0+1} \\ =-1 \end{gathered}$$