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Q. 58

Expert-verifiedFound in: Page 108

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 $47\u201360.$

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

Delta-epsilon proof is,

Whenever $0<|x-2|<\delta $, we also have $\left|\frac{{x}^{2}-3x+2}{x-2}-1\right|<\u03f5$.

We are given,

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

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

Consider that $\in >0$, choose $\delta =\u03f5$.

For all *x* with $0<|x-2|<\delta $, we also have $\left|\frac{{x}^{2}-3x+2}{x-2}-1\right|<\u03f5$.

$\left|\frac{{x}^{2}-3x+2}{x-2}-1\right|=\left|\frac{{x}^{2}-3x+2-1(x-2)}{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{(x-2{)}^{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 $\left|\frac{{x}^{2}-3x+2}{x-2}-1\right|<\u03f5$.

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