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 $x , 0 < | \mathrm { x } - \mathrm { c } | < \delta \Rightarrow | \mathrm { f } ( \mathrm { x } ) - \mathrm { L } | < \varepsilon$.
-$$f ( x ) = \frac { x ^ { 2 } + - 16 x + 60 } { x + - 6 } , c = 6 , \varepsilon = 0.02$$
A) $\mathrm { L } = 0 ; \delta = 0.02$
B) $L = - 20 ; \delta = 0.03$
C) $\mathrm { L } = - 16 ; \mathrm { \delta } = 0.03$
D) $\mathrm { L } = - 4 ; \mathrm { \delta } = 0.02$

Quizplus · Accepted Answer

The function simplifies to $f(x) = x - 10$ for $x \neq 6$, so the limit as $x$ approaches 6 is $L = -4$. The correct $\delta$ that satisfies the condition is 0.02.

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