Solutions to Dr. Karel's assignment, Calculus II class

Chapter 1, 1.7 Exercises (4,12,15,21)

4. Factor the following expression

> factor( x^2 + 3*x + 2 );

> factor( x^2 + 3.0*x + 2.0);

12. Plot and estimate the limit/value

> plot(sin(x)/x, x=-1..1);

From the graph it appears that the limit equals one.

Limit(sin(x)/x, x=0);value(``);

15. Computer exact and floating-point values

Floating Point:

> evalf(sin(Pi/4));

Exact:

> sin(Pi/4);value(``);

21. Expand then factor the following expression

> expand(exp(2*x) - 1);

> factor((exp(x))^2 - 1);

Chapter 10. Exercises 10.7 (3,4)

3. plot the following sequence and determine the limit as n->infinity using maple

> a:=n->n!/200^n;disp:=seq([n,a(n)],n=0..50):;n:='n':

(Notice the ':' just before the ';'. This ':' causes maple to hide the output of the command preceding it)

> plot([disp],style=point,axes=boxed);

For small values, the denominator 200^n has much greater value than the numerator. This is why the sequence appears to converge to zero above.

From the nature of n! and 200^n, we know that both approach infinity as n becomes very large. However, the numerator would approach

infinity more rapidly than the denominator for large values of n. Since the rate of growth is larger for the numerator than the denominator for large values of n,

the limit as n->infinity would seem to be infinity. This is supported by the results below.

> Limit(a(n),n=infinity);limit(a(n),n=infinity);

4. plot the following sequence and determine the limit as n->infinity using maple

> a:=n->(2*n^3 - 6*n^2 + 15)/ (n^4 + 18*n^3 -6);disp:=seq([n,a(n)],n=0..100):;

> plot([disp],axes=boxed, style=point);

By examining the numerator and denominator, we see that 2n^3 and n^4 are the largest terms in the numerator and denominator respectively.

To guess the behavior as n->infinity, divide the numerator by the denominator resulting in 2/n. As n->infinity, 2/n goes to zero as supported

by the graph above. The limiting value of the sequence in this case is zero as n->infintiy.

> Limit(a(n),n=infinity);limit(a(n),n=infinity);

>