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3. Given that [tex]\(\lim _{x \rightarrow 4}(2x - 2) = 6\)[/tex], illustrate this definition by finding the largest values of [tex]\(\delta\)[/tex] that correspond to:

[tex]\[
\begin{array}{ll}
c = 0.5 & \delta \leq \square \\
c = 0.1 & \delta \leq \square \\
c = 0.05 & \delta \leq \square \\
\end{array}
\][/tex]

Sagot :

To illustrate the definition given in the problem, we need to find the largest values of [tex]\(\delta\)[/tex] that correspond to the given values of [tex]\(c\)[/tex] using the limit [tex]\(\lim_{x \rightarrow 4}(2x-2)=6\)[/tex].

Given the limit [tex]\(\lim_{x \rightarrow 4}(2x-2)=6\)[/tex], our goal is to find [tex]\(\delta\)[/tex] such that:
[tex]\[ |2x - 2 - 6| < c \][/tex]
for each given value of [tex]\(c\)[/tex].

Let's tackle each [tex]\(c\)[/tex] value step by step:

### Step-by-step Solution:

#### For [tex]\(c = 0.5\)[/tex]:
We start by solving the inequality:
[tex]\[ |2x - 8| < 0.5 \][/tex]

This absolute value inequality can be split into two inequalities:
[tex]\[ -0.5 < 2x - 8 < 0.5 \][/tex]

Now solve for [tex]\(x\)[/tex]:
[tex]\[ -0.5 < 2x - 8 \rightarrow 7.5 < 2x \rightarrow \frac{7.5}{2} < x \rightarrow 3.75 < x \][/tex]
[tex]\[ 2x - 8 < 0.5 \rightarrow 2x < 8.5 \rightarrow x < \frac{8.5}{2} \rightarrow x < 4.25 \][/tex]

Combining these, we get:
[tex]\[ 3.75 < x < 4.25 \][/tex]

Next, we determine [tex]\(\delta\)[/tex] as the maximum distance from [tex]\(x = 4\)[/tex] within this interval:
[tex]\[ \delta = \min(|4 - 3.75|, |4 - 4.25|) = \min(0.25, 0.25) = 0.25 \][/tex]

Thus, for [tex]\(c = 0.5\)[/tex]:
[tex]\[ \delta \leq 0.25 \][/tex]

#### For [tex]\(c = 0.1\)[/tex]:
Now apply the same steps to the inequality:
[tex]\[ |2x - 8| < 0.1 \][/tex]

Split into two inequalities:
[tex]\[ -0.1 < 2x - 8 < 0.1 \][/tex]

Solve for [tex]\(x\)[/tex]:
[tex]\[ -0.1 < 2x - 8 \rightarrow 7.9 < 2x \rightarrow \frac{7.9}{2} < x \rightarrow 3.95 < x \][/tex]
[tex]\[ 2x - 8 < 0.1 \rightarrow 2x < 8.1 \rightarrow x < \frac{8.1}{2} \rightarrow x < 4.05 \][/tex]

Combining these, we get:
[tex]\[ 3.95 < x < 4.05 \][/tex]

Determine [tex]\(\delta\)[/tex] as the maximum distance from [tex]\(x = 4\)[/tex]:
[tex]\[ \delta = \min(|4 - 3.95|, |4 - 4.05|) = \min(0.05, 0.05) = 0.05 \][/tex]

Thus, for [tex]\(c = 0.1\)[/tex]:
[tex]\[ \delta \leq 0.05 \][/tex]

#### For [tex]\(c = 0.05\)[/tex]:
Finally, for the inequality:
[tex]\[ |2x - 8| < 0.05 \][/tex]

Split into two inequalities:
[tex]\[ -0.05 < 2x - 8 < 0.05 \][/tex]

Solve for [tex]\(x\)[/tex]:
[tex]\[ -0.05 < 2x - 8 \rightarrow 7.95 < 2x \rightarrow \frac{7.95}{2} < x \rightarrow 3.975 < x \][/tex]
[tex]\[ 2x - 8 < 0.05 \rightarrow 2x < 8.05 \rightarrow x < \frac{8.05}{2} \rightarrow x < 4.025 \][/tex]

Combining these, we get:
[tex]\[ 3.975 < x < 4.025 \][/tex]

Determine [tex]\(\delta\)[/tex] as the maximum distance from [tex]\(x = 4\)[/tex]:
[tex]\[ \delta = \min(|4 - 3.975|, |4 - 4.025|) = \min(0.025, 0.025) = 0.025 \][/tex]

Thus, for [tex]\(c = 0.05\)[/tex]:
[tex]\[ \delta \leq 0.025 \][/tex]

### Summary:
Combining all the results, we get:
[tex]\[ \begin{array}{ll} c = 0.5 & \delta \leq 0.25 \\ c = 0.1 & \delta \leq 0.05 \\ c = 0.05 & \delta \leq 0.025 \\ \end{array} \][/tex]