A function $\mathrm { f } ( \mathrm { x } )$, a point $\mathrm { c }$, the limit of $\mathrm { f } ( \mathrm { x } )$ as $x$ approaches $\mathrm { c }$, and a positive number $\varepsilon$ is given. Find a number $\delta 0$ such that for all $x , 0

Question

A function $\mathrm { f } ( \mathrm { x } )$, a point $\mathrm { c }$, the limit of $\mathrm { f } ( \mathrm { x } )$ as $x$ approaches $\mathrm { c }$, and a positive number $\varepsilon$ is given. Find a number $\delta > 0$ such that for all $x , 0 < | \mathrm { x } - \mathrm { c } | < \delta \Rightarrow | \mathrm { f } ( \mathrm { x } ) - \mathrm { L } | < \varepsilon$.
-$\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { \mathrm { x } } , \mathrm { L } = \frac { 1 } { 2 } , \mathrm { c } = 2$, and $\varepsilon = 0.3$
A) $\delta = 0.75$
B) $\delta = 0.375$
C) $\delta = 3$
D) $\delta = 10$

Quizplus · Accepted Answer

Given $f(x) = \frac{1}{x}$, $L = \frac{1}{2}$, $c = 2$, and $\varepsilon = 0.3$, we need to find $\delta$ such that if $0 < |x - 2| < \delta$, then $|\frac{1}{x} - \frac{1}{2}| < 0.3$. Solving the inequality $|\frac{1}{x} - \frac{1}{2}| < 0.3$ leads to finding the range of $x$ that satisfies this condition, which corresponds to a $\delta$ that ensures $x$ is within this range when $x$ is close to $c = 2$. The correct $\delta$ ensures the function's value is within $\varepsilon = 0.3$ of $L = \frac{1}{2}$ when $x$ is within $\delta$ of $c = 2$, and $\delta = 0.75$ satisfies this condition.

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