Give an $$ \varepsilon-\delta $$ proof of the limit fact. $$ \lim _{x \rightarrow 0}(3 x+4)=4 $$ Let $$ \varepsilon0 $$ be given. Choose the correct answer below. A. Choose $$ \delta=3 \varepsilon $$. Then for all $$ x $$ satisfying $$ 0

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Give an $$ \varepsilon-\delta $$ proof of the limit fact. $$ \lim _{x \rightarrow 0}(3 x+4)=4 $$ Let $$ \varepsilon>0 $$ be given. Choose the correct answer below. A. Choose $$ \delta=3 \varepsilon $$. Then for all $$ x $$ satisfying $$ 0<|x-0|<\delta $$, the inequality $$ |(3 x+4)-4|=|3 x|=3|x|<\frac{\delta}{3}=\varepsilon $$ holds. B. Choose $$ \delta=\frac{\varepsilon}{4} $$. Then for all $$ x $$ satisfying $$ 0<|x-0|<\delta $$, the inequality $$ |(3 x+4)-3 x|=|4 x|=4|x|<4 \delta=\varepsilon $$ holds. O. Choose $$ \delta=\varepsilon $$. Then for all $$ x $$ satisfying $$ 0<|x-0|<\delta $$, the inequality $$ |(3 x+4)-4|=|3 x|<\delta=\varepsilon $$ holds. D. Choose $$ \delta=\frac{\varepsilon}{3} $$. Then for all $$ x $$ satisfying $$ 0<|x-0|<\delta $$, the inequality $$ |(3 x+4)-4|=|3 x|=3|x|<3 \delta=\varepsilon $$ holds.

UpStudy AI · Accepted Answer

To prove that limₓ→0 (3x + 4) = 4 using an ε–δ argument, we start with the definition of a limit: For every ε > 0, there exists a δ > 0 such that if 0 < |x − 0| < δ, then |(3x + 4) − 4| < ε. Notice that |(3x + 4) − 4| = |3x| = 3|x|. We need 3|x| < ε. A natural way to ensure this is to require |x| < ε/3. Therefore, if we choose δ = ε/3 then whenever 0 < |x| < δ, we have 3|x| < 3(ε/3) = ε. This completes the proof, and thus the correct answer is: D. Choose δ = ε/3. Then for all x satisfying 0 < |x − 0| < δ, the inequality |(3x + 4) − 4| = |3x| = 3|x| < 3δ = ε holds. Choose δ = ε/3. Then for all x satisfying 0 < |x| < δ, the inequality |(3x + 4) − 4| = 3|x| < ε holds.

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