Proof that $f(x)=ax+b$ is continuous

Question

I'm looking to prove $f(x)=ax+b$ is continuous using the epsilon-delta method. Now I know it’s already been proven to be uniformly continuous and proving that means it has to be continuous anyway, but I’m still interested in how one would prove basic continuity. I would like to use the epsilon-delta method, and am having a little difficulty understanding it. So far I have,

$f(x)=ax+b$ for some $\epsilon>0$, there exists a $\delta$>0 s.t. $|x-c|<\delta$ if $|f(x)-f(c)|<\epsilon$

which means $|a(x-c)|<\epsilon$

Not really sure where to take the proof from here and would appreciate any advice, this is my first post also so apologies if the latex is wrong or anything! Again specifying that the reason I think this question hasnt been answered here is I'm not looking for uniform continuity.

Answer 1

To show, that $f:\mathbb{R}\to\mathbb{R}$, $x\mapsto ax+b$ is continuous, we have to show, that $\forall\epsilon>0\,\exists\delta>0:\forall x\in\mathbb{R}:|x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon$

Simply said, we have to find a $\delta$ for every $\epsilon$.

Keep in mind, that we have $|x-y|<\delta$ as an assumption.

You can find your $\delta$ like this. First of all:

$|f(x)-f(y)|=|ax+b-(ay+b)|=|ax+b-ay-b|=|a(x-y)|=|a||x-y|$

We know, that $|x-y|<\delta$ so we can stipulate like this:

$|a||x-y|<|a|\delta$

We want to find for every $\epsilon$ a $\delta$ such that, when $|x-y|<\delta$ we have $|f(x)-f(y)|<\epsilon$, this means that $\delta$ can be dependent from $\epsilon$.

We choose $\delta:=\frac{\epsilon}{|a|}$ for $a\neq 0$. For $a=0$ we have to find a seperate $\delta$. But this is easy, since if $a=0$, we just get $0$ and we can choose $\delta=1$ for example. It does not matter.

Note, that this is all a scatch and not the proof yet. We just searched our $\delta$.

Now for the proof:

Let $\epsilon >0$ be arbitrary. If $a=0$ choose $\delta=1$ and we get, that $|f(x)-f(y)|=|0\cdot x+b-0\cdot y-b|=0<\epsilon$.

For $a\neq 0$ choose $\delta=\frac{\epsilon}{|a|}>0$.

Then:

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

And we are done.

Answer 2

Let $e>0$ be given. Let $d=\frac{e}{2(|a|+1)}$. Then, $|x-c|<d$ implies $$ |f(x)-f(c)|=|a|\cdot|x-c|\leq(1+|a|)d<e $$ whether $a=0$ or not. Can you complete the argument now?

Answer 3

Your statement about continuity is really wrong.

In order to prove that $f$ is continuous at $c$, you need

for every $\varepsilon>0$, there exists $\delta>0$ such that, for every $x$ with $|x-c|<\delta$, it is true that $|f(x)-f(c)|<\varepsilon$.

Can you spot the difference with what you stated?

In order to prove continuity for your function, let $\varepsilon>0$ be given. Then we want to solve the inequality $|f(x)-f(c)|<\varepsilon$ and see whether the solution set has the above property. Since $$ |f(x)-f(c)|=|ax+b-ac-b|=|a||x-c| $$ we see that

1. if $a=0$, then every $\delta>0$ fits;
2. if $a\ne0$, the inequality $|a||x-c|<\varepsilon$ is equivalent to $$|x-c|<\frac{\varepsilon}{|a|},$$ so we can take $\delta=\varepsilon/|a|$.