# 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!

### Example:

\frac{(x^{2} − 1)}{(x − 1)}

Let's work it out for x=1:

\frac{(1^{2 }− 1)}{(1 − 1)} = \frac{(1 − 1)}{(1 − 1)} = \frac{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 | \frac{(x^{2} − 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 \frac{(x^{2}−1)}{(x−1)} gets **close to 2**

We are now faced with an interesting situation:

- When x=1 we don't know the answer (it is
**indeterminate**) - But we can see that it is
**going to be 2**

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 \frac{(x^{2}−1)}{(x−1)} as x approaches 1 is** 2**

And it is written in symbols as:

*lim*

**x→1**\frac{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 But we |

## 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"

## 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 (that looks good)

Example 2: 3.8 − 4 = −0.2 (**negatively** close?)

So how do we deal with the negatives? We don't care about positive or negative, we just want to know how far ... which is the absolute value.

"How Close" = |a−b|

Example 1: |4.01−4| = 0.01

Example 2: |3.8−4| = 0.2

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) = \frac{(x^{2}−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**:

δ | that |x−a| must be smaller than | |

ε | that |f(x)−L| must be smaller than |

*Note: those two Greek letters (δ is "delta" and ε is "epsilon") are*

so often used we get the phrase "delta-epsilon"

And we have:

|f(x)−L|<ε when |x−a|<δ

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

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

- it is true for any ε>0
- δ exists, and is >0
- x is
**not equal to**a, meaning 0<|x−a|

And this is what we get:

For any ε>0, there is a δ>0 so that |f(x)−L|<ε when 0<|x−a|<δ

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

But in essence it says something simple:

* f(x) gets close to L* when

*x gets close to a*## How to Use it in a Proof

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

From: | To: | |

0<|x−a|<δ | |f(x)−L|<ε |

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

How do we find such a formula?

Guess and Test!

That's right, we can:

- Play around till we find a formula that
**might**work **Test**to see if that formula does work

## Example: Let's try to show that

*lim*

**x→3**2x+4 = 10

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 we go from:

0<|x−3|<δ

to

|(2x+4)−10|<ε

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

So we can now guess that **δ=ε/2** might work

### Step 2: **Test** to see if that formula works.

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

Let's see ...

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

DONE!

We have seen then that given ε we can find a δ, so it is true that:

For any ε, there is a δ so that |f(x)−L|<ε when 0<|x−a|<δ

And we have proved that

*lim*

**x→3**2x+4 = 10

## Conclusion

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