Provide an appropriate response.
-Given $\varepsilon > 0$ , find an interval $\mathrm { I } = ( 3 - \delta , 3 ) , \delta > 0$ , such that if $x$ lies in $\mathrm { I }$ , then $\sqrt { 3 - x } < \varepsilon$ . What limit is being verified and what is its value?

Question

Provide an appropriate response.
-Given $\varepsilon > 0$, find an interval $\mathrm { I } = ( 3 - \delta , 3 ) , \delta > 0$, such that if $x$ lies in $\mathrm { I }$, then $\sqrt { 3 - x } < \varepsilon$. What limit is being verified and what is its value?
A) $\lim _ { x \rightarrow 8 ^ { - } } \sqrt { 3 - x } = 0$
B) $\lim _ { x \rightarrow 8 ^ { - } } \sqrt { x } = 3$
C) $\lim _ { x \rightarrow 8 ^ { + } } \sqrt { 3 - x } = 0$
D) $\lim _ { x \rightarrow \theta ^ { - } } \sqrt { 3 - x } = 0$

Quizplus · Accepted Answer

The limit being verified is $\lim _ { x \rightarrow 3 ^ { - } } \sqrt { 3 - x } = 0$, but due to a typographical error, none of the options accurately represent this. Option A is the closest, despite the incorrect approach to 8 instead of 3.

Provide an Appropriate Response ε>0\varepsilon > 0ε>0 , Find an Interval I=(3−δ,3),δ>0\mathrm { I } = ( 3 - \delta , 3 ) , \delta > 0I=(3−δ,3),δ>0

Provide an Appropriate Response $\varepsilon > 0$ , Find an Interval $\mathrm { I } = ( 3 - \delta , 3 ) , \delta > 0$