Expert: Sherman D. - 3/25/2005

Question
Hi,
I have just purchased the book,Calculus for Dummies,and I found that by limits, it doesn't teach this type of problem: Find the limit as x approaches infinity of [(2^x)/x]. Can you please explain to me how to do these types of problems.(Another example may be: find the limit as x approaches infinity of (x^2)/(2^x).)  

Answer
(2^x)/x

x < -infinity
x > +infinity

y > 0
y < 0

if "x" were zero, the problem would be undefined.
if you had n^x, n^x would never be zero.

if "x" were postive
(2^1)/1 = (2/1) = 2
(2^2)/2 = (4/2) = 2
(2^3)/3 = (8/3) = (2 and 2/3)
(2^4)/4 = (16/4) = 4
(2^5)/5 = (32/5) = (6 and 2/5)
(2^6)/6 = (64/6) = (10 and 2/3)
(2^+infinity)/+infinity = +infinity

if "x" were zero
(2^0)/0 = 1/0 = Undefined

x were negative numbers
(2^-1)/(-1) = (1/2)/(-1)=(1/2)/(-1/1)=(1/2)*(1/-1)=(-1/2)
(2^-1)/(-2) = (-1/4)
(2^-3)/(-3) = (-1/24)
(2^-4)/(-4) = (-1/64)
(2^-5)/(-5) = (-1/160)
(2^-infinity)/(-infinity) = -infinity

-----------------------------------------------------------

(x^2)/(2^x)

x <= 0
-infinity > y > 0

0 <= x < +infinity
0 <= y <= (9/8)
(9/8) <= y

All i can say is just to plug in values for "x" and see what you get.

x = positive numbers
(1^2)/(2^1) = (1/2)
(2^2)/(2^2) = (4/4) = 1
(3^2)/(2^3) = (9/8) = (1 and (1/8))
(4^2)/(2^4) = (16/16) = 1
(5^2)/(2^5) = (25/32)
(6^2)/(2^6) = (36/64) = (9/16)
(7^2)/(2^7) = (49/128)

x = zero
(0^2)/(2^0) = 0/1 = 0

x = negative numbers
((-1)^2)/(2^(-1)) = (1/(1/2)) = (1/1)*(2/1) = 2
((-2)^2)/(2^(-2)) = 4 * 4 = 16
((-3)^2)/(2^(-3)) = 9 * 8 = 72
((-4)^2)/(2^(-4)) = 16 * 16 = 256
((-infinity)^2)/(2^(-infinity)) = +infinity.

if you want that in integrals, just let me know.

By the way, to get a visual, go to www.calculator.com, click on Graphing, and type in (2^x)/x for the first problem, click enter on the keyboard, then click Reset when you are done, then type in (x^2)/(2^x) for the second problem, click Enter on the keyboard.