Which grows faster, (x-4)^3 or

1. Dec 14, 2009

IntegrateMe

Which grows faster as x tends to infinity, (x-4)3 or 2x3 + ln(x)?

Steps:

1) lim (x-4)3/[2x3 + ln(x)]
2) The limit is infinity/infinity as x tends to infinity, so we can use L'H rule.
3) Applying L'H rule we get [3x(x-4)2]/(6x3 + 1)
4) The limit is again infinity/infinity as the above function tends to infinity, so we apply L'H again.
5) Applying L'H rule we get [3(x-4)(3x-4)]/18x, which yet again yields infinity/infinity as x tends to infinity, so, we apply L'H rule again.
6) Applying L'H rule we get (18x-48)/18 which is equal to infinity/18 = infinity.

Solution:

(x-4)3 grows faster because the limit approaches infinity. I know i did something wrong because the answer is that they grow at the same rate. Can anyone point out my mistake(s)? Thanks!

2. Dec 14, 2009

Dick

Your expression in step 5) is not correct. Shouldn't there be an x^2 in the denominator? Can you fix it?

3. Dec 14, 2009

IntegrateMe

I feel extremely stupid right now.

So, using x^2 i would get [3(x-4)(3x-4)]/18x2
Then using L'H rule i would get: (18x-48)/18x
Then using the rule again: 18/18 = 1

Since we got a number solution, they grow at the same rate.

Wow, haha.

I have a question though, when it asks me "which grows faster," does it matter if i place (x-4)^3 over 2x^3 + ln(x) or if i place 2x^3 + ln(x) over (x-4)^3, and why?

4. Dec 14, 2009

oinkbanana

5. Dec 14, 2009

IntegrateMe

@oinkbanana, they grow at the same rate, looking at the graph + table you should see that.

We can't base our predictions on just a table and graph, which is why i went through the entire process above, so i could PROVE that one grew faster than the other, or they grew at the same rate, in which the latter turned out to be true.

6. Dec 14, 2009

Dick

It depends on what you mean by 'grow at the same rate'. If you get your limit right, then you'll figure out that the limit is actually 1/2. Which means the inverse limit is 2. Does that mean they grow at a 'different rate'?

7. Dec 14, 2009

IntegrateMe

& I have no idea what you're talking about.

8. Dec 15, 2009

Mute

Just a minor note, your calculation is still a bit off. You got the right behaviour, but you should have found the limit to tend to 1/2, not 1.

As for which one to take as the numerator - it doesn't matter. If you get a number, they grow at the same rate, if you get zero, it means the denominator grows faster, and if you get infinity it means the numerator grows faster.

Also, a handy trick for these sorts of things: while L'Hopital is the essence behind determining which functions grow faster than others, it's often a cumbersome calculation to do. What you could do for a problem like this is note that any power of x greater than 1 will grow faster than the logarithm. Hence, 2x^3 + ln(x) -> 2x^3 as x gets large. Since x beats 4 as x gets large, (x-4)^3 -> x^3, so (x-4)^3/(2x^3 + ln(x)) -> x^3/(2x^3) = 1/2 as x gets large.

Of course, if you're asked to do the problem using L'Hopital then there's really no way around doing it that way, I guess!

9. Dec 15, 2009

Dick

As mute said, i) your l'Hopital is off, the limit is 1/2. BTW just after step 1) you could have dropped the 1/x you get from the ln(x). It just goes to zero. And if lim x->infinity f(x)/g(x)=C where C is a nonzero constant, most people would say they 'grow at the same rate'.

10. Dec 15, 2009

Mentallic

Why are you confining this to only powers of x greater than 1? Don't all positive powers of x grow faster than a logarithm, including fractional powers less than 1?

What exactly does growing at the same rate mean? I understand the difference between the growth of exponentials -> polynomials -> logarithms and such, but why is the question the OP has shown considered as growing at the same rate?
One grows twice as fast as the other. Is this the same rate?

11. Dec 15, 2009

Dick

Depends on what you mean by 'rate of growth'. That's why I asked. They both grow as cubics for large x. Yes, with different coefficients, but they are both cubics.

12. Dec 15, 2009

Mentallic

Ahh ok thanks

And just before this thread goes to rest, I want clarification on the topic with a question which is too short to deserve a new thread:

Logic tells me that $e^x$ and $2^x$ grow at different rates since youre multiplying continuously by a larger number with e, therefore it should be infinitely greater at infinite. Is my logic correct?

13. Dec 15, 2009

Mute

Yes, any positive power of x will grow faster than a logarithm. It was a tired mistake on my part to say only powers > 1.

14. Dec 15, 2009

Dick

lim 2^x/e^x=0 as x->infinity, sure. Again, it would be a good idea to define what you mean by 'different rates'.