I have finals in Calculus tomorrow. Will someone help me with a few problems on my final review?? PLEASE HELP!

Question

find the limit or diverges
an= (4+9n)/(5+4n)
choices: a. (9/4) b. -29 c. (4/5) d. Diverges

find the limit or diverges
an = ln(7n - 6) - ln(5n + 7)
choices: ln (5/7) b. ln 2 c. ln (7/5) d. Diverges

find the limit or diverges
(1+ (7/n))^n
choices: e b. 1 c. e^7 d. diverges

find the sum of the series
x
"E"
n=0
(9/8n)
choices: 8 b. (72/7) c. 1 d. (9/7)

Answer 1

Hello,

1) divide the top and bottom by n, giving (4/n +9)/(5/n+4) let n approach infinity and we have (0+9)/(0+4) or 9/4 or a)

2) the difference of the logs is the log of the quotient so we have;

ln[(7n-6)/(5n+7)] now divide the numerator and denominator by n giving us ln[(7 -6/n)/(5 +7/n)] and let n approach infinity and we have ln(7/5) or c)

3) let n approach infinity and we have (1+0) raised to a large number but 1 raised to a large number is still 1 so the answer is 1 or b)

4) I don't understand the question here

Hope this helps!!

Answer 2

(4+9n)/(5+4n), towards infinity, essentially tends towards (9n)/(4n), so 9/4 - answer a.

ln(7n-6) - ln(5n+7) = ln((7n-6)/(5n+7)), tending towards ln((7n)/(5n)) = ln(7/5) - answer c.

(1+(7/n))^n = exp(ln(1+7/n))^n = exp(ln(1+7/n)*n)
ln(1+7/n), for small 7/n (n infinite), is 7/n
So we get exp(7/n*n) = exp(7) = e^7 - answer c

I don't quite get the last question...

Answer 3

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Answer 4

9/4
7/5
1
can't tell what last one is about. I see no series.

good luck

Answer 5

I can't bcuz I have never had an exam