Given $\lim _ { x \rightarrow - 3 } ( 2 - 3 x ) = 11$ find a $\delta 0$ in terms of an $\varepsilon 0$ so that if $0

Question

Given $\lim _ { x \rightarrow - 3 } ( 2 - 3 x ) = 11$ find a $\delta > 0$ in terms of an $\varepsilon > 0$ so that if $0 < | x - c | < \delta$ then $| f ( x ) - L | < \varepsilon$ A) $E$ B) $\varepsilon / 3$ C) $\varepsilon / 2$ D) None of these

Quizplus · Accepted Answer

The limit expression simplifies to $2 - 3(-3) = 11$. For $|f(x) - L| = |2 - 3x - 11| = |3x + 9|$, we have $|3(x + 3)| < \varepsilon$. Thus, $|x + 3| < \varepsilon / 3$, so $\delta = \varepsilon / 3$.

Given lim⁡x→−3(2−3x)=11\lim _ { x \rightarrow - 3 } ( 2 - 3 x ) = 11limx→−3​(2−3x)=11

Given $\lim _ { x \rightarrow - 3 } ( 2 - 3 x ) = 11$