Limits (Formal Definition)

Please read Introduction to Limits first

Approaching ...

Sometimes you can't work something out directly ... but you can see what it should be as you get closer and closer! We call that the limit of a function.

For example, what is the value of (x2-1)/(x-1) when x=1? (Notice in the graph there is actually a "hole" at that point).

Let us substitute "1" for "x" and see the result:

(12-1)/(1-1) = (1-1)/(1-1) = 0/0

But 0/0 is "indeterminate", meaning we can't determine its value, so instead of trying to work it out for x=1 let's try approaching it closer and closer:

x (x2-1)/(x-1)
0.5 1.50000
0.9 1.90000
0.99 1.99000
0.999 1.99900
0.9999 1.99990
0.99999 1.99999
... ...

Now we can see that as x gets close to 1, then (x2-1)/(x-1) gets close to 2, so we say:

The limit of (x2-1)/(x-1) as x approaches 1 is 2

And it is written in symbols as:

We don't know the value at x=1 (it is indeterminate), but the limit has a value of 2

More Formal

But you can't just say a limit equals some value because it looked like it was going to. We need a more formal definition.

So let's start with the general idea

From English to Mathematics

Let's say it in English first:

"f(x) gets close to some limit as x gets close to some value"

If we call the Limit "L", and the value that x gets close to "a" we can say

"f(x) gets close to L as x gets close to a"


Calculating "Close"

Now, what is a mathematical way of saying "close" ... could we subtract one value from the other?

Example 1: 4.01 - 4 = 0.01
Example 2: 3.8 - 4 = -0.2

Hmmm ... negatively close? That doesn't work ... we really need to say "I don't care about positive or negative, I just want to know how far" The solution is to use the absolute value.

"How Close" = |a-b|

Example 1: |4.01-4| = 0.01
Example 2: |3.8-4| = 0.2

And if |a-b| is small we know we are close, so we write:

"|f(x)-L| is small when |x-a| is small"

And this animation shows you what happens with the function

f(x) = (x2 - 1) / (x-1)

  • as x approaches a=1,
  • f(x) approaches L=2


  • |f(x)-2| is small
  • when |x-1| is small.

Delta and Epsilon

But "small" is still English and not "Mathematical-ish".

Let's choose two values to be smaller than:

that |x-a| must be smaller than
that |f(x)-L| must be smaller than

(Note: Those two greek letters, δ is "delta" and ε is "epsilon", are often
used for this, leading to the phrase "delta-epsilon")

And we have:

"|f(x)-L|<when |x-a|<"

That actually says it! So if you understand that you understand limits ...

... but to be absolutely precise we need to add these conditions:

1) 2) 3)
it is true for any >0 exists, and is >0 x not equal to a means 0<|x-a|

And this is what we get:

"for any>0, there is a >0 so that |f(x)-L|<when 0<|x-a|<"

That is the formal definition. It actually looks pretty scary, doesn't it!

But in essence it still says something simple: when x gets close to a then f(x) gets close to L.

How to Use it in a Proof

To use this definition in a proof, we want to go

From:   To:
0<|x-a|< |f(x)-L|<

This usually means finding a formula for (in terms of ) that works.

How do we find such a formula?

Guess and Test!

That's right, you can:

  1. Play around till you find a formula that might work
  2. Test to see if that formula works.

Example: Let's try to show that

Using the letters we talked about above:

  • The value that x approaches, "a", is 3
  • The Limit "L" is 10

So we want to know:

How do we go from:
0<|x-3|< to |(2x+4)-10|<

Step 1: Play around till you find a formula that might work

Start with: |(2x+4)-10|<
Simplify: |2x-6|<
Move 2 outside: 2|x-3|<
Move 2 across: |x-3|</2

So we can now guess that =/2 might work

Step 2: Test to see if that formula works.

So, can we get from 0<|x-3|< to |(2x+4)-10|< ... ?

Let's see ...

Start with: 0<|x-3|<
Replace: 0<|x-3|</2
Move 2 across: 0<2|x-3|<
Move 2 inside: 0<|2x-6|<
Replace "-6" with "+4-10" 0<|(2x+4)-10|<

Yes! We can go from 0<|x-3|< to |(2x+4)-10|<by choosing =/2


We have seen then that if we choose a we can find a , so it is true that:

"for any, there is a so that |f(x)-L|<when 0<|x-a|<"

And we have proved that


That was a fairly simple proof, but it hopefully explains the strange "there is a ... " wording, and it does show you a good way of approaching these kind of proofs.