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The logic of $\epsilon$-$\delta$ proofs.
    \begin{enumerate}
        \item Consider the fact
        \[ \lim_{x \rightarrow 4} 2x + 1 = 9 \]
        \begin{enumerate}
        \item Write out the definition of this statement using the $\epsilon$-$\delta$ definition of the limit.
        \item Take the expression $|2x+1 - 9|$ and simplify it so it has a factor of $|x-4|$ in it.
        \item If I want to have $|2x + 1 - 9| = 0.01$, what value do I need to pick for $|x-4|$? Notice that if I want $|2x+1-9| < 0.01$, I can do this by taking $|x-4|$ less than this value. 
        \item If I want to have $|2x + 1 - 9| = 0.001$, what value do I need to pick for $|x-4|$? Notice that if I want $|2x+1-9| < 0.001$, I can do this by taking $|x-4|$ less than this value. \item If I want to have $|2x + 1 - 9| = \epsilon$, what value do I need to pick for $|x-4|$? Notice that if I want $|2x+1-9| < \epsilon$, I can do this by taking $|x-4|$ less than this value. 
        \item Now, set $\delta$ to be the value you calculated in the last step. Do the calculations from the previous step in reverse to show that if $|x-4| < \delta$, then $|2x+1-9| < \epsilon$. Idea: If $|x-4| < \delta$, then we can plug that into the rewritten version of $|2x + 1 - 9|$ to get the bound that we want. 
        \end{enumerate}
        \item Now, consider the following limit.
        \[ \lim_{x \rightarrow 3} \frac{1}{x} = \frac{1}{3} \]
        \begin{enumerate}
            \item Write out the $\epsilon$-$\delta$ definition of this limit.
            \item Take the expression $\left|\frac{1}{x} - \frac{1}{3}\right|$ and rewrite it so that it has a factor of $|x-3|$.
            \item This problem is more complicated than the previous one because the extra factor attached to the $|x-3|$ is not a constant, but depends on $x$. But we can handle this. We know that we want $x$ to be close to $3$ in this limit. If $|x-3| \leq 1$, then what are the largest and smallest possible values of $x$?
            \item With those largest and smallest values of $x$, what are the largest and smallest values of this extra factor?
            \item What these last two steps have shown is that if $|x-3| < 1$, then this extra factor has an upper bound. So, as long as we make sure $\delta < 1$, that upper bound works whenever $|x-3| < \delta$. Assume that we have this, and replace the extra factor by that upper bound to get an expression of the form
            \[ \left| \frac{1}{x} - \frac{1}{3} \right| < C |x-3| \] where $C$ is the bound you found in the previous part. 
            \item This problem now looks a lot like the first example. Use the same techniques to figure out what we would need to make $\delta$ in terms of $\epsilon$ to make this definition work again.
            \item Before we go through the problem with defining $\delta$ and moving forward, we need to take into account the fact that we needed $\delta < 1$ for the non-constant factor to be bounded. Thus, we need both $\delta < 1$ and $\delta <$ some bound from the previous part. Write an expression using $\min$ in order to accomplish both of these. 
            \item Now we can pick $\delta$ to be less than this value. Work though the calculations in the reverse order to get that \[ \left| \frac{1}{x} - \frac{1}{3} \right| < \epsilon. \] Make sure you see where each of the two assumptions on $\delta$ come into play to give this result.
        \end{enumerate}
        \end{enumerate}

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