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\markright{Chapter 2: Worksheet 3}
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\Bf{\Star\ Chapter 2: Worksheet 3  \hfill Jack K. Cohen \hfill Colorado School of Mines}

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\Cb{Function Composition, Limits, Continuity}
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\noindent \Bf{Suggested Problems}
Section 2.4: 2, 6, 8, 22, 26
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\begin{enumerate}
\item
\noindent Recognizing complicated functions as compositions of simpler functions will be important later in this course.  For example if $h(x) = \sin(\cos x)$, we may introduce $u = \cos x$ and write $h$ in the simpler form $h(x) = \sin u$.  This is equivalent to writing $h(x) = f(g(x))$, where $u = g(x) = \cos x$ and $f(x) = \sin x$.  Introduce a useful decomposition for each of the following functions, $h(x)$ using both the $u$ notation and the $f, g$ notation.
	\begin{enumerate}
	\item $\sin x^3$ Note: this notation is common, but perhaps misleading---remember that it means $\sin (x^3)$, \Em{not} the function below in part (b).
	\item $\sin^3 x$  Note: this common notation for powers of trigonometric functions is misleading!  It often helps to rewrite such expressions in unambiguous notation as $(\sin x)^3$.
	\item $3^{\sin x}$
	\item $\sin(x^3 + 3)$
	\item $(x^4 + x^2 + 1)^7$
	\end{enumerate}

\item Extend the above idea to 3-fold decompositions using $u, v$ notation for the intermediate functions.  Example: For $h(x) = \sin(\cos(\tan x))$ introduce $u = \tan x$ to get $h(x) = \sin(\cos u)$ and then $v = \cos u$ to finally get $h(x) = \sin v$.
	\begin{enumerate}
	\item $\tan(\ln(x + 1))$
	\item $2^{\sin x^2}$
	\item $\sin(\sin(\sin x))$
	\end{enumerate}

\item Investigate what happens if we repeatedly compose $\sin x$ with itself.  That is, we
form the sequence $\sin x, \sin (\sin x), \sin (\sin (\sin x)), \ldots$. 	\begin{enumerate}
	\item Find the limit of this sequence when one puts $x = 0$ in each term (this is easy).
	\item What about the limit when $x = \pi / 2$?   As a start, compute
	    $\sin(\sin(\sin \pi / 2))$. (Or, look at the \Tt{Plot}s of the sucessive  composed functions).
	\end{enumerate}
 Notice that the triple composition of $\sin$ with itself is not the same thing as either of the functions discussed in problem 1, parts (a), (b)---all three are different!  If you have any doubt of this, \Tt{Plot} each of them.

\item \Star\ You have noticed that many limits can be obtained by just evaluating the given function at the given point.  State precisely a condition under which this is valid.   Which of the following limits can be evaluated in this simple way?  Justify your answer.
	\begin{enumerate}
	\item $\lim_{x \rightarrow 5}$ {\Large $\frac{x + 5}{x^4 + x^2 + 1}$}
	\item $\lim_{x \rightarrow 5}${ \Large $\frac{x^2 - 25}{x - 5}$}
	\item $\lim_{x \rightarrow 5} \sqrt{x^6 - 5}$
	\end{enumerate}

\item Here are some limits that will play a key role in developing important calculus results.  A firmly logical derivation of them is a few weeks away.  Meanwhile, employ one of your many rough and ready methods to get the right anwers to engineering accuracy.
	\begin{enumerate}
	\item $\lim_{x \rightarrow 0}${ \Large $\frac{\sin x}{x }$}
	\item $\lim_{x \rightarrow 0}${ \Large $\frac{1 - \cos x}{x }$}
	\item $\lim_{x \rightarrow 0} (1 + x)^{1/x}$
	\end{enumerate}


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