Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
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]
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]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.