# The Definition of the Limit

Recall that the limit is defined like this.

We say that the limit of a function $$f$$ at $$x_0$$ is $$L$$ if, given any positive number $$\epsilon$$, we can choose a positive number $$\delta$$ small enough that $\vert x-x_0\vert < \delta\ \textit{implies}\ \vert f(x)-L\vert < \epsilon.$

Below you can see a diagram where we have marked a value for $$x_0$$ using a dotted vertical line. It seems likely that our limit for $$f$$ at $$x_0$$ is $$L=f(x_0)$$. There are two horizontal lines marking $$L+\epsilon$$ and $$L-\epsilon$$. The red box shows the area containing all values of $$f(x)$$ with $$\vert x-x_0\vert < \delta$$. If we can choose $$\delta$$ small enough to make the box lie between the horizontal lines, then that is a $$\delta$$ that satisfies the definition.

Epsilon Plot
Fix some point $$x_0$$.
Choose $$\epsilon$$:
Then choose $$\delta$$ so that the red box fits between the horizontal lines. In other words, all the possible values of $$f(x)$$ are within epsilon of $$L$$.

## Questions

1. Given $$x_0=1.8$$ and $$\epsilon=0.1$$, give one value of delta that forces $$\vert f(x)-L\vert < \epsilon$$.
2. Again with $$x_0=1.8$$ and $$\epsilon=0.1$$, give a second value of delta that forces $$\vert f(x)-L\vert < \epsilon$$.
3. Changing $$x_0$$ to $$2.2$$ and leaving $$\epsilon=0.1$$, give a value of delta that forces $$\vert f(x)-L\vert < \epsilon$$.
4. Changing $$x_0$$ to $$1.0$$ and leaving $$\epsilon=0.1$$, give a value of delta that forces $$\vert f(x)-L\vert < \epsilon$$.