## What is a limit?

Limits are the basis for almost everything in Calculus. By the time you get to derivatives, you'll be doing limits and you won't even know it.

To become familiar with what a limit is, let's consider the function

$$f(x) = \frac{x^2-1}{x-1}.$$

What is the value of this function at $x=1$?

The function $f(x)$ is undefined at $x=1$. But the function is defined at 0.9 and 1.1:

$$f(0.9) = 1.9$$$$f(1.1)=2.1$$

In fact, the function is defined for any $x=1\pm\delta$, where $\delta \neq 0$.

In this lesson, we will be considering the case when $\delta$ is very close, but not equal to 0.

Try dragging the black point to $x=1$ in the above graph, and you will notice that the point disappears.

However, notice that it "looks like" $$f(1)=2.$$

So, we can say that

$$\lim_{x\to 1} f(x) = 2.$$

The english for this is "As $x$ approaches 1, $f(x)$ approaches 2."

Another possible interpretation is "The limit of $f(x)$ as $x$ approaches 1 is 2."

This does not mean that $f(1)=2$, but rather that if $x$ is very close to $1$, then $f(x)$ is very close to $2$.

Notice that the graphed function is linear. What linear function does it look like?

The function covers *almost* all points covered by the function $x+1$.

We can confirm this by performing the following operation:

$$\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1$$

**Important!** This kind of math is only allowed in limits. Notice how we divided the top and bottom of the fraction by $x-1$ - we only did this knowing that x is very close to, but not equal to 1.

Now try this: $$\lim_{x\to 2} \frac{x^2-5x+6}{x-2}.$$

The answer is $-1$.

As we said before, it is okay to divide by non-constants in limits. So, we can say

$$\lim_{x\to 2} \frac{x^2-5x+6}{x-2} = \lim_{x\to 2} x-3.$$

We can just plug in $x=2$ here to find

$$\lim_{x\to 2} \frac{x^2-5x+6}{x-2} = -1.$$