Finding Limit: $$\frac{8^x}{x^x}$$

[anj16]

Homework Statement

$$\lim_{x\rightarrow \infty} \frac{8^x}{x^x}$$

Homework Equations

??

The Attempt at a Solution

let $$y=\lim_{x\rightarrow \infty} \frac{8^x}{x^x}$$
$$ln y=\lim_{x\rightarrow \infty} x ln\frac{8}{x}$$

and tried l'hopital's rule but because of -∞/0 I am not able to apply that.

I tried to switch x for 1/u so that my limit goes to zero but that didn't help either.

Thank You

 

[Simon Bridge]

Good latex ... but try it this way:
$$y=\lim_{x\rightarrow \infty} \frac{8^x}{x^x}$$

First get a feel for the expression:
Looking at it, as x gets very big, which gets bigger faster: $8^x$ or $x^x$ ?

 

[anj16]

Good latex ... but try it this way:
$$y=\lim_{x\rightarrow \infty} \frac{8^x}{x^x}$$

First get a feel for the expression:
Looking at it, as x gets very big, which gets bigger faster: $8^x$ or $x^x$ ?

$x^x$

Which makes the limit go to zero. Correct?
I agree to the above logic but is there no other way of showing it? The thing is if this were on a test or an exam my teacher will definitely not give me 5/5 by just saying that. So I am trying to find a way of showing that the limit goes to zero instead of just stating the above reasoning.

By the way what is latex??

 

[That Neuron]

Okay, I would suggest is a power tower derivation in conjunction with L'Hospital's rule.

If we derive 8^x we get ln8 (e^x). If we derive x^x we get (e^x)^lnx, if we take u as equal to e^x and say that u approaches infinity as x approaches infinity we end up with lim u → ∞ ln8 (u/(u^ln2u), if we use L'Hospital's rule again with respect to u, we get ln8 (ln²x X x^ln2x/ x²lnx. Which doesn't really help much. My guess would be to say that since u ^ln2u is always larger than u, so it approaches 0.

 

[anj16]

Okay, I would suggest is a power tower derivation in conjunction with L'Hospital's rule.

If we derive 8^x we get ln8 (e^x). If we derive x^x we get (e^x)^lnx, if we take u as equal to e^x and say that u approaches infinity as x approaches infinity we end up with lim u → ∞ ln8 (u/(u^ln2u), if we use L'Hospital's rule again with respect to u, we get ln8 (ln²x X x^ln2x/ x²lnx. Which doesn't really help much. My guess would be to say that since u ^ln2u is always larger than u, so it approaches 0.

Thanks for the suggestion but I don't see how the denominator is bigger that the numerator after you differentiate and substitute. Maybe I am just reading it wrong.

But assuming you are correct (which you most probably are) how is this anymore convincing that what simon posted.

 

[Simon Bridge]

$x^x$

Which makes the limit go to zero. Correct?
I agree to the above logic but is there no other way of showing it? The thing is if this were on a test or an exam my teacher will definitely not give me 5/5 by just saying that. So I am trying to find a way of showing that the limit goes to zero instead of just stating the above reasoning.

Well you would have to do something to put that reasoning on a mathematical footing, yes.

Note: is $x^a/y^a = (x/y)^a$ any help?

By the way what is latex??

It is the way that equations are marked up.
See how the way I wrote the equations was nice-looking ... that is what latex does. It is everything inside the "$$" marks in my last post.

 

[anj16]

Note: is $x^a/y^a = (x/y)^a$ any help?

And that is how I took the natural log and reached the conclusion of -∞/0, but I believe l'hopital's rule cannot be applied in this situation, correct? Or are you referring to something else?

 

[Dick]

And that is how I took the natural log and reached the conclusion of -∞/0, but I believe l'hopital's rule cannot be applied in this situation, correct? Or are you referring to something else?

Just take the log of the expression. Use the rules of logs. You don't even have to use l'Hopitals. Just look at it. Show what you got. And besides, the original problem in the form (8/x)^x has the form 0^∞. That's not even indeterminant.