Please help me!

[reaper8060]

1. Lim{x->0}[(x^x^x)/x]
...
2. Lim{x->+∞}[(e.x^(x+1))/((x+1)^x) - x]
...
Sorry i can't use math type.!
Thanks everyone!

 

[daon2]

I assume your first limit is as x approaches zero from the right.

First, find

\(\displaystyle \displaystyle \lim_{x\to 0^+} x^x\)

Then proceed as follows


\(\displaystyle \dfrac{x^{\displaystyle x^x}}{x} = x^{\displaystyle x^x-1} = e^{\displaystyle (x^x-1)\ln(x)}\)

 

[reaper8060]

I assume your first limit is as x approaches zero from the right.

First, find

\(\displaystyle \displaystyle \lim_{x\to 0^+} x^x\)

Then proceed as follows


\(\displaystyle \dfrac{x^{\displaystyle x^x}}{x} = x^{\displaystyle x^x-1} = e^{\displaystyle (x^x-1)\ln(x)}\)

Sorry i forgot!. Yes,x approaches zero from the right
please present more detailed, i don't understand!

How to solve lim(lnx.(x^x-1)) ???

I proceed:

lim{x->0+}(x^x) =1 => lim{x->0+}[(x^x^x)/x] =(x^1)/x =1 ???

I'm not sure. Is it correct???

 

[daon2]

Sorry i forgot!. Yes,x approaches zero from the right
please present more detailed, i don't understand!

How to solve lim(lnx.(x^x-1)) ???

I proceed:

lim{x->0+}(x^x) =1 => lim{x->0+}[(x^x^x)/x] =(x^1)/x =1 ???

I'm not sure. Is it correct???

You cannot do that. If the base of the exponent were independent of the variable that is allowed, not in this case.

This limit is harder than I anticipated actually. Is this some kind of "extra credit" question? Or for an advanced class?

edit:
I gave up trying to find an elementary solution. I had one posted here but I found an error, it might be able to be modified (it shows, intuitively at least, the limit is 1) so I'll try again after work. Either way, interesting problem. My thought is to use the squeeze theorem. A good lower bound is given by \(\displaystyle x\le x^{x^x}\). I haven't yet found a convenient upper bound.

 

[reaper8060]

You cannot do that. If the base of the exponent were independent of the variable that is allowed, not in this case.

This limit is harder than I anticipated actually. Is this some kind of "extra credit" question? Or for an advanced class?

Edit
I gave up trying to find an elementary solution. I had one posted here but I found an error, it might be able to be modified (it shows, intuitively at least, the limit is 1) so I'll try again after work. Either way, interesting problem. My thought is to use the squeeze theorem. A good lower bound is given by \(\displaystyle x\le x^{x^x}\). I haven't yet found a convenient upper bound.

:lol:
Thank you for caring!
Indeed, finding a function which can be applied squeeze theorem isn't easy
i also tried, but it's no result

"Yeh, I'm learning in an engineering programs! it's slightly harder than other schools. During the first year of university, study program is heavy on theory. "


(And, sorry if my english is not good)
Thanks you!

 

[daon2]

:lol:
Thank you for caring!
Indeed, finding a function which can be applied squeeze theorem isn't easy
i also tried, but it's no result

"Yeh, I'm learning in an engineering programs! it's slightly harder than other schools. During the first year of university, study program is heavy on theory. "


(And, sorry if my english is not good)
Thanks you!

I found one.

\(\displaystyle x \le x^{x^x} \le xe^{\sqrt{x}}\)

I'll leave it to you to find an interval \(\displaystyle (0, a)\) for which this is true (it might be hard to pinpoint, but using the graph here: link might help you). Then

\(\displaystyle 1 \le \dfrac{x^{x^x}}{x} \le e^{\sqrt{x}}\)

Taking limits, \(\displaystyle 1 \le L \le 1\)