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 \(\vert f(x)-L\vert < \epsilon\).