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\begin{enumerate}
\item Evaluate $\lim_{x\rightarrow 1} x$.
\item Why can't we evaluate  \[ \lim_{x\rightarrow 1} \frac{x^2 - x}{x-1} \] by direct substitution? What do we get?
\item We want to try to compute this limit anyway. Factor the numerator. Since we are \textbf{not} looking at the value at $x=1$, we can cancel a factor. What do we get now?
\item Define
\[ f(x) = \begin{cases}
\frac{x^2 - x}{x-1} & x \neq 1 \\
a & x=1
\end{cases} \]
What value should be assigned to $a$ in to make $f$ a continuous function on $\mathbb{R}$?
\item Use the same factoring procedure to try to evaluate the following limits. Note that direct substitution always gives the same `answer' initially. 
\[ \lim_{x\rightarrow 2} \frac{x^2-4}{(x-2)^3} \qquad \lim_{x\rightarrow 3} \frac{(x-3)^2}{x^2-9} \qquad \lim_{x \rightarrow -2} \frac{x^2 + 3x + 2}{x^2 + 4x + 4} \]
\item What does that say about problems where that same answer shows up by direct substitution?
\end{enumerate}

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