Mathwizurd.com is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

# What is a Limit?

A limit can be difficult to define, however it essentially is the value a function approaches as the x approaches a number (or infinity).

For example,

Here, obviously, when y = 2x + 5, and x gets closer and closer to 3, y gets closer and closer to 11.

The important thing about limits is that they should work either way: the left-hand limit (coming from the left) and the right-hand limit (coming from the right).

For example,

When x -> 0- (from the left), it's 0, but when x -> 0+ (from the right), it's 1. So, there are two possible limits, meaning the limit DOES NOT EXIST.

# Formal Definition of a Limit

Let's say you want to prove to someone that a curve has a limit. So, you say that given any bounds (like y = 1 and y = -1 below), you will give a point that satisfies the epsilon; in this case it's 1.

If you want to prove that the limit approaching a real number exists, you have to say that for all epsilon, there exists a value, delta, such that if x is less than delta units away from the limit, the f(x) will be less than epsilon away from the limit.

Below, the limit approaches infinity, so you have to do something different. Given any epsilon, there exists a delta such that if x > delta, then |f(x)| < epsilon.

# Proof of Limit to infinity

Here is an example proof for the one above: f(x) = 1/x, and we want to find the limit of f(x) as x -> infinity. This will be in two phases: the discovery phase, where we discover what we want del ta to be, and the proof, where we go backwards.

### Discovery Phase

```x > 0 (given)
|f(x) - 0| < epsilon #|f(x) - Limit| < epsilon
1/x < epsilon (x > 0, don't need abs sign)
x > 1/eps (flip signs)
```

### Proof

```Now, we let delta = 1/eps
x > delta, so x > 1/eps
1/x < eps
|1/x| < eps
|f(x)| < eps```

So, the second part was just the first part reversed.

# Proof with of Limit to a Number

Now, the proof is kind of different, the steps are the same, but now if x is < delta units away from c, f(x) is less than epsilon units from L, the limit. It can be written like, if |x-c| < delta, then |f(x) - L| < epsilon.

So, let's prove the limit from the top.

### Discovery Phase

```|2x + 5 - 11| < Eps
|2x - 6| < Eps
2|x-3| < Eps
|x-3| < Eps/2
```

### Proof

```Let Delta = Eps/2
|x-3| < Delta = Eps/2
|x-3| < Eps/2
|2x - 6| < Eps
|2x + 5 - 11| < Eps
|f(x) - 11| < Eps
```
David Witten