I just want to make sure that is this idea is true for all $y=mx+b$ cases: for a line of slope $m$, then $\delta=\frac{\epsilon}{\left|m\right|}$ (or "could at most equal").
In other words, the steeper the line, the smaller $\delta$ is to $\epsilon$
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2If you think about the geometric meaning of the $\epsilon$ and the $\delta$, you should be able to convince yourself that this is true: $\epsilon$ sets up a “horizontal band”; if the line is very steep, then you will end up outside that band very “quickly” (not very far from the point $x_0$) which means you need $\delta$ to be small so as not to stray to far. – Arturo Magidin Feb 07 '20 at 06:01
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Assuming the slope $m \in \mathbb{R}$ is non-zero and $b\in \mathbb{R}$, your postulate follows directly from the definition since you given $\varepsilon >0$ and some $x_0\in \mathbb{R}$ want to find $\delta>0$ s.t. for all $x\in \mathbb{R}$ where $|x-x_0|<\delta$ it holds that $$ |(mx+b)-(mx_0+b)|<\varepsilon. $$ Evaluting the lhs we see that $$ |(mx+b)-(mx_0+b)|=|mx+b-mx_0-b|=|mx+mx_0|=|m(x-x_0)|=|m||x-x_0|<\varepsilon. $$ Dividing by $|m|$ obtains your result.
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