Given $\lim _ { x \rightarrow 2 } ( x + 3 ) = 5$ find a $\delta 0$ in terms of an $\varepsilon 0$ so that if $0

Question

Given $\lim _ { x \rightarrow 2 } ( x + 3 ) = 5$ find a $\delta > 0$ in terms of an $\varepsilon > 0$ so that if $0 < | x - c | < \delta$ then $| f ( x ) - L | < \varepsilon$ A) /2 B) /3 C) 1 D) None of these.

Quizplus · Accepted Answer

The limit statement $\lim _ { x \rightarrow 2 } ( x + 3 ) = 5$ implies that for any $\varepsilon > 0$, we need to find $\delta > 0$ such that if $0 < |x - 2| < \delta$, then $|(x + 3) - 5| < \varepsilon$. Simplifying, $|x + 3 - 5| = |x - 2| < \varepsilon$. Thus, $\delta = \varepsilon$.

Given lim⁡x→2(x+3)=5\lim _ { x \rightarrow 2 } ( x + 3 ) = 5limx→2​(x+3)=5

Given $\lim _ { x \rightarrow 2 } ( x + 3 ) = 5$