Chapter 2: Worksheet 3 ANSWERS MA 131

    1. u = x3, h(x) = sin u or g(x) = x3, f (x) = sin x
    2. u = sin x, h = u3
    3. u = sin x, h = 3u
    4. u = x3 +3, h = sin u
    5. u = (x4 + x2 +1), h = u7

    1. u = x + 1, v = ln u, h = tan v
    2. u = x2, v = sin u, h = 2v
    3. u = sin x, v = sin u, h = sin v

    1. 0
    2. 0 (see Figure 1 and 2). Note sinπ/2 = 1.0 and sin(sin(sinπ/2)) = 0.745624. Since π/2 produces the largest value, once we have the result there, it isn't hard to realize that the function goes to 0 at every x as we take more and more compositions.

    Figure: 4 compositions of sin with itself
    \begin{figure}
\epsfysize 80pt
\centerline{\epsffile{ans3p3a.eps}}\end{figure}

    Figure: 128 compositions of sin with itself
    \begin{figure}
\epsfysize 80pt
\centerline{\epsffile{ans3p3b.eps}}\end{figure}

  1. Merely evaluating the function to get the limit is justified when the function is continuous. If you don't see this important point, study the definition of continuity carefully and/or ask your instructor.
    1. Rational functions are continuous except at points where the denominator vanishes. Hence we may just ``plug in'' for this example: $\lim_{{x \rightarrow 5}}^{}$ ${\frac{{x + 5}}{{x^4 + x^2 + 1}}}$ = ${\frac{{5 + 5}}{{625 + 25 + 1}}}$ = ${\frac{{10}}{{651}}}$
    2. Here we have the favorite case in calculus where numerator and denominator both vanish at a. Because of the denominator, the function is not continuous at x = 5. However, in evaluating the limit, we do not have to consider this point, so we may employ factoring for x≠5 to get $\lim_{{x \rightarrow 5}}^{}$(x + 5) = 10.
    3. No problem, just ``plug in'': $\lim_{{x \rightarrow 5}}^{}$$\sqrt{{x^6 - 5}}$ = $\sqrt{{5^5 - 5}}$ = $\sqrt{{5 \cdot 624}}$

    1. 1 (For example, make a Table or use the Program of the Week)
    2. 0
    3. e $\approx$ 2.71828