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Let f(x) = ax+ b Let any x,y ∈ R. Let ε > 0 be abritraty

|f(x) – f(y)| = |ax+b – (ay+b)|

= |ax-ay| < ε


= |a(x-y)| < ε 


≤ |a||x-y| < ε


If |a||x-y| < ε then |x-y| < ε / |a|

Let δ = ε / |a| > 0 |f(x)-f(y) < ε for |x-y|< δ where δ = ε / |a|

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  • $\begingroup$ You should distinguish the case when $a=0$, but in that case it is obvious. $\endgroup$ – egreg Dec 10 '16 at 20:26
  • $\begingroup$ You can just proves that $f$ is Lipschitz continuous, which implies uniform continuity. $\endgroup$ – Henricus V. Dec 11 '16 at 2:48
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I agree with your answer. Here is how I would have formulated it.

We need to prove that $$ \forall \epsilon > 0 \ \exists \delta : \forall x,y \in \mathbb{R} \ |x-y| < \delta \Rightarrow |f(x) - f(y)|< \epsilon $$

Now let $\epsilon > 0$ and assume $a \not = 0$, choose $\delta = \frac{\epsilon}{|a|}$, then

$$ |x-y| < \delta = \frac{\epsilon}{|a|} \Leftrightarrow |a| |x - y | = | ax - ay| = |(ax - b) -(ay - b)| < \epsilon $$

Thus $\delta$ does not depend on $x,y$.

As pointed out in the comment under this answer, for $a = 0$, you can choose $\delta$, for example, to be $1$.

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  • $\begingroup$ For $a\ne0$, of course. For $a=0$, just take $\delta=1$ or whatever. $\endgroup$ – egreg Dec 10 '16 at 20:27
  • $\begingroup$ You are correct, Thanks for pointing it out @egreg. $\endgroup$ – Olba12 Dec 10 '16 at 20:33
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    $\begingroup$ Doesnt for a=0, any value of δ work? So could I just show the equation and say "any value works" or should I show an actual value for δ? $\endgroup$ – Aggrawal Puja Dec 10 '16 at 20:40
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    $\begingroup$ Yes, indeed, any works. But if you read the defintion, it says $\exists \delta$ which means that there is atleast one. Hence it is sufficient to show the uniform continuity for only one $\delta$ of your choosing. @AggrawalPuja $\endgroup$ – Olba12 Dec 10 '16 at 20:42

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