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 $x , 0 < | \mathrm { x } - \mathrm { c } | < \delta \Rightarrow | \mathrm { f } ( \mathrm { x } ) - \mathrm { L } | < \varepsilon$.
-$\mathrm { f } ( \mathrm { x } ) = \sqrt { 17 - \mathrm { x } } , \mathrm { L } = 4 , \mathrm { c } = 1$, and $\varepsilon = 1$
A) $\delta = - 9$
B) $\delta = 8$
C) $\delta = 12$
D) $\delta = 7$

Quizplus · Accepted Answer

The Answer of 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 $x , 0 < | \mathrm { x } - \mathrm { c } | < \delta \Rightarrow | \mathrm { f } ( \mathrm { x } ) - \mathrm { L } | < \varepsilon$.
-$\mathrm { f } ( \mathrm { x } ) = \sqrt { 17 - \mathrm { x } } , \mathrm { L } = 4 , \mathrm { c } = 1$, and $\varepsilon = 1$
A) $\delta = - 9$
B) $\delta = 8$
C) $\delta = 12$
D) $\delta = 7$

A Function f(x)\mathrm { f } ( \mathrm { x } )f(x)

A Function $\mathrm { f } ( \mathrm { x } )$