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Clearly, this problem requires applying epsilon, delta definition of uniform continuity. im having trouble finding values that will work here and hence, writing the proof in general.

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    $\begingroup$ See the answers to this question. $\endgroup$ – Dietrich Burde Dec 10 '16 at 19:23
  • $\begingroup$ What have you tried so far? What criteria must be satisfied to have uniform continuity? You're a lot more likely to get help if you've shown that you've at least made an effort. $\endgroup$ – NNN Dec 10 '16 at 19:50
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We say $f$ is uniformly continuous if

$ \forall \, \varepsilon > 0 \; \exists \, \delta > 0 \,$ such that $\forall x,y\in \mathbb{R}$ we have $|x-y|<\delta \Rightarrow |f(x) - f(y)| < \varepsilon$.

If $a=0$, then $|f(x)-f(y)| = |b-b| = 0 < \varepsilon$ for any $\varepsilon$, so any $\delta$ works.

Now let $\varepsilon>0$ and suppose $a \neq 0$. Let $\delta = \frac{\varepsilon}{|a|}$ and suppose $|x-y|<\delta$. Then

$$ |f(x) - f(y)| = |ax + b - (ay + b)| = |ax - ay| = |a||x-y| < |a|\delta = |a|\frac{\varepsilon}{|a|} = \varepsilon $$

and we're done.

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Here is what I came up with . Does this look correct?

I figured out two values that work but not sure if this proof makes sense here

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  • $\begingroup$ Only problems: $a$ might be negative, in which case you want to divide by the absolute value, and $a$ might be zero, in which case you want a slightly different approach. $\endgroup$ – Ian Dec 10 '16 at 19:52
  • $\begingroup$ Should I use the absolute value of (a) here then? instead of just a? $\endgroup$ – Aggrawal Puja Dec 10 '16 at 19:53
  • $\begingroup$ That'll handle the nonzero case yes. $\endgroup$ – Ian Dec 10 '16 at 20:58

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