Question

question 20/1 pt 5 19 detailswhen demonstrating that $lim_{x \to -1} 3x + 5 = 2$ with $varepsilon = 0.3$, which of the following $delta$-values suffices? there may be more than one correct answer, so select all that apply$square delta =0.1$$square delta =0.01$$square delta = 0.03333333333333$$square delta =0.2$

Explanation:

Step1: Recall epsilon-delta definition

We need \(|(3x+5)-2| < 0.3\) when \(0 < |x - (-1)| < \delta\), simplify the left inequality:
\(|3x + 3| < 0.3\)
\(3|x + 1| < 0.3\)

Step2: Solve for |x+1|

Divide both sides by 3:
\(|x + 1| < \frac{0.3}{3} = 0.1\)

Step3: Valid δ values

Any \(\delta \leq 0.1\) satisfies the condition, since a smaller δ will still keep \(|x+1|\) within the required bound.

Answer:

$\delta =0.1$, $\delta =0.01$, $\delta = 0.03333333333333333$