Limits (Formal Definition)

Please read Introduction to Limits first

Approaching ...

Sometimes we can't work something out directly ... but we can see what it should be as we get closer and closer!


(x2 − 1) (x − 1)

Let's work it out for x=1:

(12 − 1) (1 − 1) = (1 − 1) (1 − 1) = 0 0

Now 0/0 is a difficulty! We don't really know the value of 0/0 (it is "indeterminate"), so we need another way of answering this.

So instead of trying to work it out for x=1 let's try approaching it closer and closer:

Example Continued:

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 see that as x gets close to 1, then (x2−1) (x−1) gets close to 2

We are now faced with an interesting situation:

We want to give the answer "2" but can't, so instead mathematicians say exactly what is going on by using the special word "limit"

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

And it is written in symbols as:

limit as x goes to 1 of (x^2-1)/(x-1) = 2

So it is a special way of saying, "ignoring what happens when we get there, but as we get closer and closer the answer gets closer and closer to 2"

As a graph it looks like this:

So, in truth, we cannot say what the value at x=1 is.

But we can say that as we approach 1, the limit is 2.

 graph hole

More Formal

But instead of saying a limit equals some value because it looked like it was going to, we can have 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"

When 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"

limit idea: f(x) goes to L as x goes 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 yes
Example 2: 3.8 − 4 = −0.2 not

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" which is the absolute value.

"How Close" = |a−b|

Example 1: |4.01−4| = 0.01 yes
Example 2: |3.8−4| = 0.2 yes

And when |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 what happens with the function

f(x) = (x2−1) (x−1)

f(x) approaches L=2 as x approaches a=1,
so |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:

delta that |x−a| must be smaller than
epsilon  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|<epsilonwhen |x−a|<delta"

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 epsilon>0 deltaexists, and is >0 x not equal to a means 0<|x−a|

And this is what we get:

"for anyepsilon>0, there is a delta>0 so that |f(x)−L|<epsilonwhen 0<|x−a|<delta"

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|<deltaright arrow|f(x)−L|<epsilon

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

How do we find such a formula?

Guess and Test!

That's right, we can:

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

Example: Let's try to show that

limit as x goes to 3 of 2x+4 = 10

Using the letters we talked about above:

So we want to know:

How do we go from:

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

Start with:|(2x+4)−10|<epsilon
Move 2 outside:2|x−3|<epsilon
Move 2 across:|x−3|<epsilon/2

So we can now guess that delta=epsilon/2 might work

Step 2: Test to see if that formula works.

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

Let's see ...

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

Yes! We can go from 0<|x−3|<delta to |(2x+4)−10|<epsilonby choosing delta=epsilon/2


We have seen then that given epsilonwe can find a delta, so it is true that:

"for anyepsilon, there is a delta so that |f(x)−L|<epsilonwhen 0<|x−a|<delta"

And we have proved that

limit as x goes to 3 of 2x+4 = 10


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.