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 is the value a function or sequence approaches as the input approaches another value.

For example,

MathJax TeX Test Page $$\lim_{x \to 3} \left(2x + 5\right) = 11$$

Here, 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 approaches 0 from the left, the value approaches 0. However, when x approaches 0 from the right, it's 1. So, there are two possible limits, meaning the limit DOES NOT EXIST.

# Formal Definition of a Limit of a Sequence

A limit is formally defined as you can find always an input that will be arbitrary close to the limit. Let’s say your sequence is 1/x. If you want to be within 0.0001 of 0, x has to be greater than 10000.

MathJax TeX Test Page Definition: If $p \in \mathbb{R}^n$ and $\{x_i\}^{\infty}_{i = 1}$ is a sequence in $\mathbb{R}^n$, then
$\{x_i\}^{\infty}_{i = 1}$ converges to $p$ iff $\forall \epsilon>0, \exists k \in \mathbb{N}$ s.t. $\forall i \geq k, ||x_i - p|| < \epsilon$
This can be written as $$\lim_{i \to \infty}x_i = p$$

In other words, in order for your limit to exist, if I suggest any given epsilon, you have to respond with a k such that all sequence elements after that will be within epsilon of the limit.

# 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. The discovery phase is secret. You don’t want people seeing this part. It’s like an iPhone. Apple doesn’t want you to know the the hundreds of failed attempts, they only want you to know the final result.

### Discovery Phase

MathJax TeX Test Page In the discovery phase, we start with a guess, and we reverse engineer the necessary x. We guess that the limit approaches 0. $$||f(x_i) - 0|| < \epsilon$$ $$||\dfrac{1}{i}|| < \epsilon$$ We know that $\dfrac{1}{i}$ is positive, because $x \geq 1$. $$i > \dfrac{1}{\epsilon}$$

### Proof

MathJax TeX Test Page $$\text{Let } \epsilon > 0$$ $$\text{Let }k = \lceil 1 + \dfrac{1}{\epsilon}\rceil$$ We know from our discovery phase that $k > \dfrac{1}{\epsilon}$, where k is the minimum index of x. k must also be an integer. The equation above satisfies both properties. $$\text{Let }i \geq k$$ $$i \geq \lceil 1 + \dfrac{1}{\epsilon}\rceil > \dfrac{1}{\epsilon}$$ $$\epsilon > \dfrac{1}{i} = ||x_i - 0||$$ $$||x_i - 0|| < \epsilon$$ Here is the logic of the proof: from the last 4 lines, we can conclude $$\forall i \geq \lceil 1 + \dfrac{1}{\epsilon}, ||x_i - p|| < \epsilon$$ Going one line back, we get $$\exists k s.t. \forall i \geq \lceil 1 + \dfrac{1}{\epsilon}, ||x_i - p|| < \epsilon$$ Finally, the "Let $\epsilon > 0$" lets us say $$\forall \epsilon > 0 \exists k s.t. \forall i \geq \lceil 1 + \dfrac{1}{\epsilon}, ||x_i - p|| < \epsilon$$ That is the definition of a limit of a sequence.

## 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.

MathJax TeX Test Page The idea of a limit to a point is for any epsilon greater than 0, we can create a range of x-values ($\delta$) such that if x is in the x-range, the f(x) is in the y-range. Written out formally, it is this: $$\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. if } 0 < ||x - p||< \delta \to ||f(x) - f(p)|| < \epsilon$$ Once again, we have to do a discover phase. This means we start with our epsilon and have to find a delta.

### Discovery Phase

MathJax TeX Test Page $$|2x + 5 - 11| < \epsilon$$ $$|2x - 6| < \epsilon$$ $$|x-3|< \dfrac{\epsilon}{2}$$ Now, we have our delta.

### Proof

MathJax TeX Test Page $$\text{Let } \epsilon > 0$$ $$\text{Let } \delta = \dfrac{\epsilon}{2}$$ $$|x - 3| < \dfrac{\epsilon}{2}$$ $$2|x-3| < \epsilon$$ $$|2x - 6| < \epsilon$$ $$|2x + 5 - 11| < \epsilon$$ Therefore, we can say $\forall \epsilon > 0, \exists \delta > 0\text{ s.t. if }|x - 3| < \delta \to |f(x) - 11| < \epsilon$
David Witten