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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Indeterminate Forms and Algebraic Evaluation of Limits}
\centerline{October 3, 2017}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
Most of the limits we will need to deal with in calculus are so-called \emph{indeterminate forms}
like $0/0$, $\infty/\infty$ (and eventually also $0\cdot\infty$ and $\infty - \infty$), 
where the Limit Laws and continuity don't help immediately. For instance
$$\lim_{x\to 1} \frac{x^4 - x^3}{x^2 - 1}$$
is a $0/0$ form since the numerator and denominator of this rational function both 
tend to $0$ as $x \to 1$.  
The perhaps surprising thing is that in cases like this we can often 
\begin{enumerate}
\item[(1)] algebraically cancel a factor producing the $0$'s in the numerator and denominator, and/or
\item[(2)] rearrange the function in other ways
\end{enumerate}
to see that a limit exists and evaluate it. 
For example, the limit above can be evaluated like this.  After factoring and cancelling a
common factor of $x - 1$, we get: 
$$\frac{x^4 - x^3}{x^2 - 1} = \frac{x^3(x - 1)}{(x + 1)(x - 1)} = \frac{x^3}{x + 1}.$$
This new function is equal to the original one for all $x \ne 1$, but it is continuous at $x = 1$.  
Hence we can see
$$\lim_{x \to 1}  \frac{x^4 - x^3}{x^2 - 1} = \lim_{x\to 1} \frac{x^3}{x + 1} = \frac{1}{2}.$$
 
This method will always involve replacing the original function 
by some other function that is defined and continuous at the $x$-value where we are taking the limit, 
and then the limit can be obtained by substitution, \emph{after the replacement}.  

The following questions will ``lead you through'' some first examples.  \emph{Eventually, though, you will need 
to be able to recognize which algebraic methods apply and work through examples like this without the prompts 
from me.}  
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\noindent
{\it Questions}
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\begin{enumerate}
\item[(1)] Consider the limit 
$$\displaystyle{\lim_{x\to 4} \frac{x^3 - 64}{x^2 - 16}},$$ 
\begin{enumerate}
\item[(a)]  First explain why this is a $0/0$ form.
\item[(b)]  Next, factor the top and 
bottom.  You should find a common factor of $x - 4$ that can be cancelled.  
\item[(c)] Do this, and then note
that ``what is left'' is continuous at $x = 4$, so we can find the limit by substitution in the 
cancelled form.  What is the limit?
\end{enumerate}
\item[(2)] Next, consider this limit:
$$\lim_{x\to 3} \frac{\sqrt{x^2 + 16} - 5}{x - 3}.$$
\begin{enumerate}
\item[(a)]  Explain why this is also a $0/0$ form.
\item[(b)]  We can look at the top as $\sqrt{x^2 + 16} - \sqrt{25}$.  In cases like this, 
multiply top and bottom by the conjugate
radical $\sqrt{x^2 + 16} + \sqrt{25}$ and simplify the top.  
\item[(c)] You should now see something to cancel so that the remaining function is continuous
at $x = 3$.  Do that and determine the limit.
\end{enumerate}
\item[(3)]  Next consider this limit:
$$\lim_{x\to 5} \frac{\frac{1}{x^2} - \frac{1}{25}}{x - 5}$$
\begin{enumerate}
\item[(a)]  Why is this one an indeterminate $0/0$ form?
\item[(b)]  Put the two terms on the top over a common denominator.  Then simplify the fraction 
and find something to cancel.  
\item[(c)]  Evaluate the limit.
\end{enumerate}
\item[(4)]  Many $\infty/\infty$ indeterminate form limits can be evaluated in a similar
way.  We'll consider 
$$\lim_{x\to \pi/2} \frac{\tan(x)}{\sec(x)}$$
\begin{enumerate}
\item[(a)]  Explain why this one is an $\infty/\infty$ indeterminate form.
\item[(b)]  Rewrite in terms of $\sin(x)$ and $\cos(x)$ and simplify, cancelling something in the process.
\item[(c)]  What is left should be continuous at $x = \pi/2$, so evaluate the limit by substitution.
\end{enumerate}
\end{enumerate}
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