Kristen Robinson

EMAT 4680

21 November 2000

Assignment 23

1.) Given a choice between $1,000,000 at once today or $.01 today, $.02 tomorrow, $.04 the next day, etc. for 30 days, I would choose the latter option. The total amount received at the end of thirty days would be $10,737,418.23. This is 9,737,418.23 more than the first option of $1,000,000 at once. After 27 days, I would already have over $1,000,000 with a total of $1,342,177.27. This amount is twice as great as the amount of $671088.63 from the day before.

2.) a.) I chose to find the limit as x approaches 3 when h(x) = (x+4)/((x^2)-25). By examining the range from 2.9995 to 3.0005 increasing by .0001 each time, I found the limit as x approaches 3 to be -.4375 or -7/16.

b.) I chose to find the limit as x approaches -3 for the same h(x). By examining the range from -3.0005 to -2.9995 increasing by .0001 each time, I found the limit as x approaches

-3 to be -.0625 or -1/16.

c.) One undefined value of h(x) would be when x = 5. By examining the range from 4.9995 to 5.0005 increasing by .0001 each time, I found the limit as x approaches 5 from the left to be -9000 and the limit as x approaches 5 from the right to be 9000. Therefore, there is a vertical asymptote at x = 5.

d.) To find the limit as x approaches positive infinity, I examined the range from 1000 to 11000 increasing by 1000 each time. I found the limit as x approaches positive infinity to be 0.

e.) To find the limit as x approaches negative infinity, I examined the range from -1000 to -11000 decreasing by 1000 each time. I found the limit as x approaches negative infinity to be 0.

3.) a.) As n increases, the ratio of each pair of adjacent terms approaches the limit of 1.618033989. As n increases, the ratio of every second term approaches the limit of 2.618033989 which is the square of 1.618033989. Likewise, as n increases, the ratio of every third term approaches the limit of 4.236067977 which is (1.618033989)^3. Therefore, the ratio of every xth term approaches the limit (1.618033989)^x as n increases.

b.) Let f(0) = 5 and f(1) = 2. As n increases, the ratio of each pair of adjacent terms approaches the same limit of 1.618033989. Likewise, the ratio of every xth term approaches the limit (1.618033989)^x as n increases regardless of the value of f(0) and f(1).

4.) I have tried countless sets of numbers, positive numbers, negative numbers, decimals, square roots, extremely large numbers, etc. I have not been able to get more than 7 rows before all zeroes are generated in the calculation.