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Q. 58

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 in Exercises $47–60.$$\underset{x\to 2}{\mathrm{lim}}\frac{{x}^{2}-3x+2}{x-2}=1$

Delta-epsilon proof is,

Whenever $0<|x-2|<\delta$, we also have $\left|\frac{{x}^{2}-3x+2}{x-2}-1\right|<ϵ$.

See the step by step solution

## Step 1. Given information

We are given,

$\underset{x\to 2}{\mathrm{lim}}\frac{{x}^{2}-3x+2}{x-2}=1$

## Step 2. Writing the delta-epsilon proofs

The strategy is to write delta-epsilon proofs for the given limit statement.

Consider that $\in >0$, choose $\delta =ϵ$.

For all x with $0<|x-2|<\delta$, we also have $\left|\frac{{x}^{2}-3x+2}{x-2}-1\right|<ϵ$.

$\left|\frac{{x}^{2}-3x+2}{x-2}-1\right|=\left|\frac{{x}^{2}-3x+2-1\left(x-2\right)}{x-2}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{{x}^{2}-3x+2-x+2}{x-2}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{{x}^{2}-4x+4}{x-2}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{\left(x-2{\right)}^{2}}{x-2}\right|\phantom{\rule{0ex}{0ex}}=|x-2|\phantom{\rule{0ex}{0ex}}<\delta \phantom{\rule{0ex}{0ex}}=\in$

So, whenever $0<|x-2|<\delta$, we also have $\left|\frac{{x}^{2}-3x+2}{x-2}-1\right|<ϵ$.