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 Math Help > Calculus > Limit > Limit Intro

## Introduction to the Limit

The limit, as x approaches c, of a function f(x), is said to be equal to L if the value of f(x) gets "closer and closer" to L as x gets "closer and closer" to c.  This page presents a mathematically rigorous explanation of this, and also introduces some special uses of limits.

First some nomenclature:

"the limit, as x approaches c, of f(x), is L",

and it means roughly,

The closer x gets to c, the closer f(x) gets to L

or to put it just a little more rigorously -- first, let me explain the idea of a limit.  It's kind of like a legal contract.  Imagine there are two people, Mr. Believer and Mr. Doubter.  Believer claims that the closer x gets to c, the closer f(x) gets to L, but Doubter tries to refute this claim.  Doubter says "Hey, when x is far away from c, f(x) is pretty close to L, but when you bring x just a little closer to c, f(x) moves away from L, so your claim is false!"

Believer responds, "Yeah, but I'm only concerned with the behavior of f when x really, really close to x.  In that case, when you bring it even closer, f(x) moves even closer to L."

At that point, Doubter hauls off and punches Believer, pretty much ending the discussion.

Clearly these two guys need a way of stating this fact that can be used objectively to settle any sort of argument like this one.

Here's the method that mathematicians have developed:

Let f be a function defined on an open interval containing c (except possibly at c) and let L be a real number.  For any given positive real number, ε, (pronounced "epsilon"), there exists a positive real number, δ, (pronounced "delta"), such that if x isn't equal to c but is within δ of c, then f(x) will be within ε of L.

Here, the colloquial expression "x is within δ of c" means |x - c| < δ, so the statement can be made more in a more symbolic way like this:

Means given any positive real number, ε, there exists a positive real number, δ, such that .

Now, if Believer and Doubter accept this definition, then all Believer has to do to show the limit is really L is this: write a formula that depends on ε and gives δ as a result.  Something like δ = ε²/1000, for example.  Then Believer writes a series of statements that prove that for the given function, and a given value of ε that whenever

0 <  |x - c| < δ (where δ comes from Believer's formula that uses ε)

it follows that

|f(x) - L| < ε

"Sin x over x" shows an example that illustrate how Believer would find δ as a function of ε.

Next, I will show you a graph of a function that illustrates the meanings of δ and ε...

### Limit of sum equals sum of limits

if limx−>a f(x)=k and limx−>a g(x)=l, prove: limx−>a f(x)+g(x)=k+l

The way to prove this from first principles is to go back to the definition of "limit".

Since the limit as x->a of f(x)=k, the definition tells us that for any positive real number, e, there exists a positive real number, d, such that 0 < |x-a| < d implies |f(x)-k)| < e. (statement 1)

Since the limit as x->a of g(x)=l, the definition tells us that for any positive real number, q, there exists a positive real number, p, such that 0 < |x-a| < p implies |g(x)-l| < q. (statement 2)

Now, to prove that the limit as x->a of f(x)+g(x)=k+l, consider any positive real number, s.  Using statements 1 and 2, let e=q=s/2.  From statements 1 and 2, we see that d and p exist with particular properties as mentioned.  Let r be the smaller of d and p.

From statement 1, 0 < |x-a| < r implies |f(x)-k|<s/2 and |g(x)-l|<s/2.  From the triangle inequality, then, |f(x)+g(x)-k-l| < s, proving the limit.

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The webmaster and author of this Math Help site is Graeme McRae.