- #1
bwpbruce
- 60
- 1
Given:
\(\displaystyle \lim_{x \to 3} \dfrac{2}{x + 3} = \dfrac{1}{3}\)
Goal:
Prove using $\epsilon-\delta$ definition of a limit.
Plan:
Let \(\displaystyle \epsilon > 0 \) be given. Find \(\displaystyle \delta > 0 \) so that
if \(\displaystyle 0 < \left|x - 3 \right| < \delta \), then \(\displaystyle \left|f(x) - \dfrac{1}{3} \right| < \epsilon \)
Solve:
\(\displaystyle \left|\dfrac{2}{x + 3} - \dfrac{1}{3} \right| = \left|\dfrac{6}{3(x + 3)} - \dfrac{x + 3}{3(x + 3)} \right| = \left| \dfrac{x + 3}{3(x + 3)} - \dfrac{6}{3(x + 3)}\right| = \left| \dfrac{x + 3 - 6}{3(x + 3)}\right|= \left| \dfrac{x -3}{3(x + 3)}\right| < \epsilon \)
\begin{align*} |x - 3| &< |3(x + 3)|\epsilon \\ x\text{ is approaching }3, \text{therefore }|x - 3| &< |3(3 + 3)|\epsilon \\ |x - 3| &< |3(6)| \\ \left|x - 3 \right| &< 18\epsilon\end{align*}
Result:
\(\displaystyle 18\epsilon = \delta \)
Conclusion:
From the statement \(\displaystyle 18\epsilon = \delta \), it is clear that for any given \(\displaystyle \epsilon\), a \(\displaystyle \delta\) exists such that if \(\displaystyle 0 < \left|x - 3 \right| < \delta \), then \(\displaystyle \left|f(x) - \dfrac{1}{3} \right| < \epsilon \)
\(\displaystyle \lim_{x \to 3} \dfrac{2}{x + 3} = \dfrac{1}{3}\)
Goal:
Prove using $\epsilon-\delta$ definition of a limit.
Plan:
Let \(\displaystyle \epsilon > 0 \) be given. Find \(\displaystyle \delta > 0 \) so that
if \(\displaystyle 0 < \left|x - 3 \right| < \delta \), then \(\displaystyle \left|f(x) - \dfrac{1}{3} \right| < \epsilon \)
Solve:
\(\displaystyle \left|\dfrac{2}{x + 3} - \dfrac{1}{3} \right| = \left|\dfrac{6}{3(x + 3)} - \dfrac{x + 3}{3(x + 3)} \right| = \left| \dfrac{x + 3}{3(x + 3)} - \dfrac{6}{3(x + 3)}\right| = \left| \dfrac{x + 3 - 6}{3(x + 3)}\right|= \left| \dfrac{x -3}{3(x + 3)}\right| < \epsilon \)
\begin{align*} |x - 3| &< |3(x + 3)|\epsilon \\ x\text{ is approaching }3, \text{therefore }|x - 3| &< |3(3 + 3)|\epsilon \\ |x - 3| &< |3(6)| \\ \left|x - 3 \right| &< 18\epsilon\end{align*}
Result:
\(\displaystyle 18\epsilon = \delta \)
Conclusion:
From the statement \(\displaystyle 18\epsilon = \delta \), it is clear that for any given \(\displaystyle \epsilon\), a \(\displaystyle \delta\) exists such that if \(\displaystyle 0 < \left|x - 3 \right| < \delta \), then \(\displaystyle \left|f(x) - \dfrac{1}{3} \right| < \epsilon \)