Questions: Sketch the interval (a, b) on the x-axis with the point c inside. Then find the largest value of δ>0 such that for all x, 0<x-c<δ implies a<x<b. a=1/8, b=1/3, c=1/4

Sketch the interval (a, b) on the x-axis with the point c inside. Then find the largest value of δ>0 such that for all x, 0<x-c<δ implies a<x<b.
a=1/8, b=1/3, c=1/4

Transcript text: Sketch the interval $(a, b)$ on the $x$-axis with the point c inside. Then find the largest value of $\delta>0$ such that for all $\mathrm{x}, 0<|\mathrm{x}-\mathrm{c}|<\delta$ implies $\mathrm{a}<\mathrm{x}<\mathrm{b}$. \[ a=\frac{1}{8}, b=\frac{1}{3}, c=\frac{1}{4} \]

Solution

Solution Steps

Step 1: Identify the interval and point

Given the interval $(a, b)$ on the $x$-axis with the point $c$ inside: \[ a = \frac{1}{8}, \quad b = \frac{1}{3}, \quad c = \frac{1}{4} \]

Step 2: Determine the distance from $c$ to the endpoints of the interval

Calculate the distance from $c$ to $a$ and $b$: \[ \delta_1 = c - a = \frac{1}{4} - \frac{1}{8} = \frac{1}{8} \] \[ \delta_2 = b - c = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \]

Step 3: Find the largest value of $\delta$

The largest value of $\delta$ is the minimum of $\delta_1$ and $\delta_2$: \[ \delta = \min\left(\frac{1}{8}, \frac{1}{12}\right) = \frac{1}{12} \]

Final Answer

The largest value of $\delta > 0$ such that for all $x$, $0 < |x - c| < \delta$ implies $a < x < b$ is: \[ \delta = \frac{1}{12} \]

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