Question

find the maximum value of \\(\delta > 0\\) that satisfies the limit claim corresponding to \\(\varepsilon = 0.3\\), that is, such that \\(0 < |x - c| < \delta\\) would imply \\(|f(x) - l| < 0.3\\).

\\\lim_{x \to 4} (5x - 1) = 19\\

maximum value of \\(\delta\\):

Explanation:

Identify the given limit and parameters

We are given the limit:

$$ \lim_{x \to 4} (5x - 1) = 19 $$

Here, the function is \( f(x) = 5x - 1 \), the limit point is \( c = 4 \), and the limit value is \( L = 19 \).
We are given \(\varepsilon = 0.3\).

Set up the epsilon inequality

Using the Epsilon-Delta Definition, we want to find the maximum \(\delta > 0\) such that:

$$ |f(x) - L| < \varepsilon $$

Substituting the given values:

$$ |(5x - 1) - 19| < 0.3 $$

Simplify the inequality

Simplify the expression inside the absolute value:

$$ |5x - 20| < 0.3 $$

Factor out the constant \(5\):

$$ 5|x - 4| < 0.3 $$

Solve for delta

Divide both sides of the inequality by \(5\):

$$ |x - 4| < \frac{0.3}{5} $$

$$ |x - 4| < 0.06 $$

This matches the form \(0 < |x - c| < \delta\), where \(c = 4\).
Thus, the maximum value of \(\delta\) is \(0.06\).

Answer:

Find the maximum value of \(\delta > 0\) that satisfies the limit claim corresponding to \(\varepsilon = 0.3\), that is, such that \(0 < |x - c| < \delta\) would imply \(|f(x) - L| < 0.3\).

$$ \lim_{x \to 4} (5x - 1) = 19 $$

Maximum value of \(\delta\): <blank>0.06</blank>