Question

What is the definition of limit in calculus?

Answer 1

There are several ways of stating the definition of the limit of a function. In order for an alternative to be acceptable it must give the same results as the other accepted definitions. Those other definitions are accepted exactly because they do give the same results.

The definition of the limit of a function given in textbooks used for Calculus I in the U.S. is some version of:

Definition
Let `f` be a function defined on some open interval containing `a` (except possibly at `a`).
Then the limit as `x` approaches `a` of `f` is `L`, written:

`color(white)"ssssssssss"` `lim_(xrarra)f(x)=L`

if and only if

for every `epsilon > 0` there is a `delta > 0` for which:

if `0 < abs(x-a) < delta`, then `abs(f(x) - L) < epsilon`.

That is the end of the definition

Comments
Tlhe following version is a bit more "wordy", but it is clearer to many.

for every `epsilon > 0` (for every positive epsilon),
there is a `delta > 0` (there is a positive delta)

for which the following is true:

if `x` is any number for which `0 < abs(x-a) < delta` is true,
then `abs(f(x) - L) < epsilon` is also true.

An acceptable rephrasing of that "if . . ., tlhen . . . " is:

If `x` is a chosen number within distance `delta` of `a` (but `x!=a` because weird stuff might happen right at `a`),

then `f(x)` is a number within distance `epsilon` of `L`

The Game

I claim that `lim_(xrarra) f(x) = L`.

What I am claiming is that:

if someone else chooses how close I need to make `f(x)` to `L` (give me a distance `epsilon`)

then I'll show that if you start with an `x` close enough to `a` (within `delta` of `a`) then you'll get an `f(x)` within `epsilon` of `L`.
Because I am only making a claim about values of `f(x)` for `x`'s chose to `a`, the `x` chosen cannot be equal to `a`.
(For nice functions this last bit about `x=a` won't matter, but not all functions are nice.)