f(x) = 2x+1, proving that it is continuous when p = 1 with 𝛿 and ε

[thethagent]

TL;DR Summary: Continuity of a function, Calculus newbie, delta, epsilon,

Greetings! I have just started studying Calculus for my engineering course, and I am already facing some problems to understand the fundamental ideas regarding the continuity of a function. I'd be very much grateful if you guys spared a minute or two to help me with this question:

f(x) = 2x+1, p = 1, prove with delta and epsilon notation that it is continuous for p

How I'd start it

1 - 𝛿 < x < 1 + 𝛿 => 3 - ε < f(x) < 3+ ε
No problems so far. I'd change f(x) for its function and then I'd have a inequality for x on both sides

1 - 𝛿 < x < 1 + 𝛿 => 1 - ε/2 < x < 1 + ε/2

From this point I'm starting to have some problems. The author of the book assumes that 1 - 𝛿 = 1 - ε/2 and 1 + 𝛿 = 1 + ε/2, but I cannot fathom why it is true. For instance, I could say that x = 3, then
0 < 3 < 6 and 2 < 3 < 9

My point with it is: x shouldn't necessarily be limited by the same two things, which is why 1 - 𝛿 = 1 - ε/2 and 1 + 𝛿 = 1 + ε/2 isn't necessarily true. Yet, the book states it, so I must be missing something here. Can you help me with it?

I appreciate anyone who has read so far; thank you so much!

 

[PeroK]

I can't really follow what you are trying to do.

In general, an epsilon-delta proof is of the form:

Let $\epsilon > 0$. Choose $\delta =$ some function of $\epsilon$ (making sure that $\delta > 0$). Then show that:
$$|x - p| < \delta \implies |f(x) - f(p)| < \epsilon$$In this case,given that $f$ is a linear function, it should be clear that it's sufficient to choose $\delta = \dfrac \epsilon 2$.

I would redo your proof using this standard approach.

 

[PeroK]

PS I don't want to give you too much help, but note that:
$$|f(x)- f(p)| = |(2x +1) - (2p +1)| = |2x - 2p| = 2|x - p|$$This is the way you should think about it.

 

[FactChecker]

A "word-smithing" note on your statement of the problem:
It wasn't until I saw @PeroK 's answer that I realized what p had to do with the problem. A more clear statement of the problem would be:
Show that f(x)=2x+1 is continuous at the point x=1.

 

[fresh_42]

Here is a little article on how to deal with the $\varepsilon$-calculus.
https://www.physicsforums.com/insights/epsilontic-limits-and-continuity/#Continuity

 

[pasmith]

My first question was "what is p?"

In fact what you are trying to do is show that f(x) = 2x + 1 is continuous at x = 1.

 

[fresh_42]

My first question was "what is p?"

In fact what you are trying to do is show that f(x) = 2x + 1 is continuous at x = 1.

And it is a very poor example where almost nothing can be learned from. It invites to confuse the roles of $\varepsilon$ and $\delta ,$ and the dependence of the location $x=1$ doesn't occur.

@thethagent, if you want to learn something, then prove that $f(x)=\begin{cases}\dfrac{1}{x}&\text{ for }x\neq 0\\0 &\text{ for }x=0\end{cases} \quad$ is continuous at $x=0.2$ and discontinuous at $x=0.$

 

[FactChecker]

From this point I'm starting to have some problems. The author of the book assumes that 1 - 𝛿 = 1 - ε/2 and 1 + 𝛿 = 1 + ε/2, but I cannot fathom why it is true.

The technique of the proof is: given an arbitrarily small $\epsilon \gt 0$, define $\delta \gt 0$ such that $|x-1|\lt\delta$ forces $|f(x)=f(1)|\lt\epsilon$.
Suppose we have an arbitrarily small $\epsilon\gt 0$. Can we define a $\delta$ that will work? The author is proposing that if $\delta$ is defined so that $1-\delta=1-\epsilon/2$ it will work. He is correct. From that definition, $\delta=\epsilon/2$. Suppose that $|x-1|\lt\delta=\epsilon/2$. Can you continue the proof from there?