Answer true or false. It can be shown that if |f(x) - L|

Name: Answer true or false. It can be shown that if |f(x) - L| 2 - 9| < $\varepsilon$ if |x - 3| < $\delta$ for arbitrarily small positive $\varepsilon$.
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Question

Answer true or false. It can be shown that if |f(x) - L| < $\varepsilon$ when 0 < |x - a| < $\delta$, then |x² - 9| < $\varepsilon$ if |x - 3| < $\delta$ for arbitrarily small positive $\varepsilon$.

Quizplus · Accepted Answer

If |f(x) - L| < F1F1F1S1F1F1F10 when 0 < |x - a| < F1F1F1S1F1F1F10, then |f(x) - L| is small when x is close to a. Since x² - 9 = (x - 3)(x + 3), if |x - 3| is small, then |x² - 9| is also small.

Answer True or False ε\varepsilonε When 0 < |X - A|

Answer True or False $\varepsilon$ When 0 < |X - A|