Monday, November 30, 2015

Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 26

Show that the statement $\displaystyle\lim\limits_{x \to 0} x^3 = 0$ 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^3-0| < \varepsilon\\

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
& \text{That is,}\\
& \phantom{x} & \text{ if } 0 < |x| < \delta \qquad \text{ then } \qquad |x^3| < \varepsilon\\
\end{aligned}
\end{equation}
$

Or, taking the cube root of both sides of the inequality $|x^3| < \varepsilon$ , we get...
$\quad \text{if } 0 < |x| < \delta \quad \text{ then } \quad |x| < \sqrt[3]{\varepsilon}$

The statement suggests that we should choose $\displaystyle \delta = \sqrt[3]{\varepsilon}$

By proving that the assumed value of $\delta = \sqrt[3]{\varepsilon}$ will fit the definition...



$
\begin{equation}
\begin{aligned}
\text{if } 0 < |x| < \delta \text{ then, }\\
& \phantom{x} &
|x^3| < \delta^3 = (\sqrt[3]{\varepsilon})^3 = \varepsilon

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \text{Thus, }\\
& \phantom{x} \quad\text{if } 0 < |x| < \delta \qquad \text{ then } \qquad |x^3| < \varepsilon\\
& \text{Therefore, by the definition of a limit}\\
& \phantom{x} \qquad \lim\limits_{x \to 0}x^3 = 0

\end{aligned}
\end{equation}
$

No comments:

Post a Comment

Why is the fact that the Americans are helping the Russians important?

In the late author Tom Clancy’s first novel, The Hunt for Red October, the assistance rendered to the Russians by the United States is impor...