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Answers without the blur. Sign up and see all textbooks for free! Q. 48

Expert-verified Found in: Page 108 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Write delta-epsilon proofs for each of the limit statements $\underset{x\to c}{\mathrm{lim}}f\left(x\right)=L$ in Exercises $47-60$.$\underset{x\to 2}{\mathrm{lim}}\left(3-4x\right)=-5$.

Delta-epsilon proof for $\underset{x\to 2}{\mathrm{lim}}\left(3-4x\right)=-5$ is, whenever localid="1648006806212" $0<\left|x-2\right|<\delta$, we also have localid="1648021669064" $\left|\left(3-4x\right)+5\right|<\epsilon$.

See the step by step solution

## Step 1. Given information

$\underset{x\to 2}{\mathrm{lim}}\left(3-4x\right)=-5$.

## Step 2. Consider ε>0, choose δ=ε4.

For all $x$ with localid="1648006934128" $0<\left|x-2\right|<\delta$,we also have localid="1648021795730" $\left|\left(3-4x\right)+5\right|<\epsilon$.

localid="1648021807616" $\left|\left(3-4x\right)+5\right|=\left|3-4x+5\right|\phantom{\rule{0ex}{0ex}}=\left|8-4x\right|\phantom{\rule{0ex}{0ex}}=4\left|2-x\right|\phantom{\rule{0ex}{0ex}}=4\left|x-2\right|\phantom{\rule{0ex}{0ex}}<4\delta \phantom{\rule{0ex}{0ex}}=4\left(\frac{\epsilon }{4}\right)\phantom{\rule{0ex}{0ex}}=\epsilon$

Therefore, whenever $0<\left|x-2\right|<\delta$, we also have localid="1648021825608" $\left|\left(3-4x\right)+5\right|<\epsilon$. ### Want to see more solutions like these? 