Why is Delta Dependent on Epsilon in the Formal Definition of a Limit?

[Bashyboy]

I am reading about the formal definition of a limit, and its corresponding proof, and there is one thing that I don't quite understand, yet. It says that delta depends on epsilon, but what I wonder is why is it not the other way around. Indeed, why does delta have dependency on epsilon?

 

[Whovian]

It's basically saying,

"For every epsilon, there is a delta such that ..."

Try wrapping your head around how this definition works first, it should be more intuitive then.

 

[thrill3rnit3]

You want that delta to work for any choice of ε > 0.

 

[AlephZero]

It says that delta depends on epsilon, but what I wonder is why is it not the other way around.

To prove the limit exists, if somebody gives you a value of $\epsilon$ (any value they like, so long as $\epsilon > 0$) you have to find a $\delta$ that satisfies the definition.

The definition of a limit says something about how the function behaves at a single point, not about how it behaves at every point in an interval. So asking "what happens in an interval of size $\delta$ around the point" and trying to find a value of $\epsilon$ might not make any sense.

For example think about the function $f(x) = 1/x$ if $x \ne 0$ and $f(x) = 0$ when $x = 0$. This is continuous at every point except $x = 0$. But if you take $x = a$ and $\delta > a$ you can't say anything about $\epsilon$ in the interval $|x-a| < \delta$, which includes points on either side of $x = 0$. Of course to prove $f(x)$ is continuous at $x = a$, you use values of $\delta$ that are small enough so the interval does NOT include $x=0$.

Indeed, why does delta have dependency on epsilon?

Work out what the definition means, for some simple functions like $f(x) = 2x$. For most functions, the smaller the value of $\epsilon$, the smaller you have to make $\delta$.

 

[SteveL27]

I am reading about the formal definition of a limit, and its corresponding proof, and there is one thing that I don't quite understand, yet. It says that delta depends on epsilon, but what I wonder is why is it not the other way around. Indeed, why does delta have dependency on epsilon?

In order to constrain the output by a given amount; you must constrain the input by an amount that depends on the contraint on the output.

 

[Fredrik]

We want the value of the function to get arbitrarily close to the limit. If we denote the limit by y, this means that for all positive numbers ε, the function must at some point in its domain have a value in the interval (y-ε,y+ε). This is why the definition starts with "For all ε>0".

Now consider the function f defined by f(x)=sin(1/x) for all x>0. (I edited this sentence after micromass' correction below).

If we only require that the function get arbitrarily close to a limit, then every number in the interval [-1,1] would be a limit of this function at 0. For example, when x approaches 0, f(x) gets arbitrarily close to 1. But we don't want a definition that makes every number in [-1,1] a limit of this function at 0. We want a definition that ensures that this function doesn't have a limit at 0. So we require not only that the function comes close to the limit, but also that it stays close to the limit.

To be more precise, we say that y is a limit of f at b, or equivalently, a limit of f(x) as x goes to b, if there's an open interval around b such that all the values of f in that interval are in the interval (y-ε,y+ε). In other words, we require that there's a δ>0 such that for all x in (b-δ,b+δ), f(x) is in (y-ε,y+ε).

If you understand this, it should be pretty obvious that given a small ε, you have to choose a small δ to ensure that the last requirement is satisfied. This is why δ depends on ε.

 

[micromass]

Now consider the function f defined by f(x)=(sin x)/x for all x>0.

That function plotted isn't sin(x)/x
Did you rather mean sin(1/x)?? That would make sense...

 

[micromass]

You want that delta to work for any choice of ε > 0.

No, that's the entire point. You don't want that delta to work for any epsilon. You want a delta to work for each epsilon.

 

[Fredrik]

That function plotted isn't sin(x)/x
Did you rather mean sin(1/x)?? That would make sense...

Oops. Yes, of course. I will edit my post. Thanks.