Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 25

Show that the statement $\displaystyle\lim\limits_{x \to 0} x^2 = 0$ is correct using the $\varepsilon$, $\delta$ definition of limit.

Based from the defintion,


$\begin{equation} \begin{aligned} \phantom{x} \text{if } & 0 < |x - a| < \delta \qquad \text{ then } \qquad |f(x) - L| < \varepsilon\\ \phantom{x} \text{if } & 0 < |x-0| < \delta \qquad \text{ then } \qquad |x^2 - 0| < \varepsilon\\ \end{aligned} \end{equation}$




$\begin{equation} \begin{aligned} & \text{That is,}\\ & \phantom{x} & \text{ if } 0 < |x| < \delta \qquad \text{ then } \qquad |x^2| < \varepsilon \\ \end{aligned} \end{equation}$

Or, taking the square root of both sides of the inequality $|x^2| < \varepsilon$, we get
$\quad \text{ if } 0 < |x| < \delta \qquad \text{ then } \qquad |x| < \sqrt{\varepsilon}$

The statement suggests that we should choose $\displaystyle \delta = \sqrt{\varepsilon}$

By proving that the assumed value of $\delta = \sqrt{\varepsilon}$ will fit the definition...



$\begin{equation} \begin{aligned} \text{if }\quad 0 < |x| < \delta \quad \text{ then,}\\ \quad |x^2| < \delta^2 = (\sqrt{\varepsilon})^2 = \varepsilon\\ \end{aligned} \end{equation}$




$\begin{equation} \begin{aligned} & \text{Thus, }\\ & \phantom{x} \quad\text{if } 0 < |x-0| < \delta \qquad \text{ then } \qquad |x^2-0| < \varepsilon\\ & \text{Therefore, by the definition of a limit}\\ & \phantom{x} \quad \lim\limits_{x \to 0} x^2 = 0 \end{aligned} \end{equation}$