Prove: $\displaystyle\lim_{x \to 5}3x+2=17$
General format $\displaystyle\lim_{x \to a}f(x)=L$ (included to verify various parameters)
What needs to be proved: $\forall\epsilon>0, \exists\delta>0$, such that $0<|x-a|<\delta|\implies|f(x)-L|<\epsilon$
Let $\epsilon>0$, $\epsilon$ be arbitrary, pick $\delta=\color{red}{\epsilon/3}\implies3\delta=\epsilon$ (Picked later)
Hypothesis: $\exists\delta>0$ (I am afraid of saying for every $x \in\mathbb{R}$), there exists a $x$ such that $0<|x-5|<\delta\implies|3x+2-17|<\epsilon$
Proof:
Now, $|3x+2-7|=|3x-15|=3|x-5|$
$|x-5|<\delta\implies3|x-5|<3\delta=\epsilon/3\implies|x-5|<\epsilon$ (Q.E.D)
This proof is based on https://www.youtube.com/watch?v=dXr6aoJ1nVI
Remark:
I redo to see if I can write a proof that follows logical order.
This proof somehow doesn't seem to me to spell out that there exists such an $x$, while approaching $a$ will make the distance between $|f(x)-L|<\epsilon$. Did I prove it correctly?
Also with this kind of proof, I find that you usually pick a $\delta$ in such a way that eventually will give you a $\epsilon$. Also this kind of excercise is easy to do because $|x-a|$ can be written as $N|f(x)-a|$, with $N$ being a constant factor out to make $|f(x)-L|=|x-a|$
Do you think doing the proof as shown in the video with a double inequality is better?
Best Answer
The point of an $\epsilon,\delta$ proof of a limit is to find $\delta$ given $\epsilon$, because we are after all proving that for any $\epsilon$, there exists a $\delta$ ...
With this in mind, let's start the proof.
Proof: Let $\epsilon>0$. We must find $\delta>0$ such that $$|x-5|<\delta\Rightarrow |3x+2-17|<\epsilon.$$ Keep in mind that our choice of $\delta$ depends only on the value of $\epsilon$. We may notice that $|3x+2-17|<\epsilon$ is equivalent to the condition $-\epsilon<3x-15<\epsilon$, so $$-\epsilon/3<x-5<\epsilon/3.$$ However, our desired value $\delta$ is defined to satisfy $-\delta<x-5<\delta$. Thus, $\delta=\epsilon/3$ is a perfect choice!
The proof is now complete, because we were given any value $\epsilon>0$, and from that we were able to prove the existence of (in other words, find) a value $\delta>0$ such that $$|x-a|<\delta\Rightarrow |f(x)-L|<\epsilon.$$