Expert: Sherry Wallin - 4/4/2010

Question
Hello, I will post the entire question and then explain my dilemma.

Suppose that f, g are positive functions. If lim as x-> infinity f(x)/g(x) = 0, we say that f grows more slowly that g or that g grows more rapidly than f. This may be indicated by writing:
f(x) << g(x). Arrange the following functions in order of rate of growth from least rapidly growing to most rapidly growing:

x
ln(x)
square root(x^3 + x^2 + x + 1)
e^x
ln(ln(x))
ln(e^x^2 + e^x + 1)
xln(x)
2^x^2
square root(10^x + 5^x +1)
x^x

For example I found out that for one of them I can do this:

2^x^2 >> x^x because,

ln(2^x^2) >> ln(x^2) therefore
x^2ln(2) >> xln(x)

I was thinking about trying to graph all of the equations and go from there but I am not certain that will yield anything. I was wondering if you can offer any advice as to how to go about solving this problem.

Answer
Hi Alex~
   In your example you said: 2^x^2 >> x^x because,

ln(2^x^2) >> ln(x^2) therefore
x^2ln(2) >> xln(x)

but ln(2^x^2)= x^2ln(2)and ln(x^2)= 2lnx, not xlnx. It is true though that x^2ln(2)>>2lnx

Your idea about graphing is a good technique to see what is going on with the functions and I highly recommend it. Because your 'rate of growth' is defined for the limit as x -> infinity you are more interested in what happens way out on the x axis. Since this is Calculus II I don't think your instructor expects you to prove what you arrive at. There are ways to put a bound on a function which is what you seem to be trying to do. What you might do is substitute a very 'large' value for x in each function and see what you have. For example what if x = 100?

x = 100
ln(100)= 4.61
square root(100^3 + 100^2 + 100 + 1)= 1005
e^100 = 2.688 x 10^43
ln(ln(100))= 1.528
ln(e^100^2 + e^100 + 1)= overflow error
100 ln100 = 461
2^10000 = overflow error
square root(10^100 + 5^100 +1)= overflow error
100^100 = overflow error

So x = 100 is 'too large' but what you can say is that you have the first 6 in the least to largest in growth rate:
ln(ln(100))= 1.528, ln(100)= 4.61, x = 100, 100 ln100 = 461,
square root(100^3+100^2+100+1)= 1005, e^100 = 2.688 x 10^43
that is:   ln(ln(x))<<ln(x)<<x<<xlnx<<square root(x^3+x^2+x+1)<<e^x

Now try another number that is large but not as large as x = 100, maybe x = 10 and see what happens to the remaining functions

ln(e^10^2 + e^10 + 1)= 100
2^10^2 = 1.267 x 10^30
square root(10^10 + 5^10 +1)= 100048.8
10^10 = 10,000,000,000

Now from this you can safely say that:
ln(ln(x))<<ln(x)<<x<<xlnx<<square root(x^3+x^2+x+1)<<e^x<<ln(e^x^2 + e^x + 1)
<<square root(10^x + 5^x +1)<<x^x<<2^x^2

Math Prof

Graph them and see what you think