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#19
Jun1307, 01:05 AM

P: 74




#20
Jun1307, 01:15 AM

P: 1,705




#21
Jun1307, 01:18 AM

P: 390

As you can see, log x's derivative is always less than x's derivative that is to say that x goes to infinity faster than log(x) would. (Equivalent to what L'Hopital would tell you I guess) That's one way to look at it but morphism would've won me over with that elegant application of the squeeze theorem. 


#22
Jul908, 05:06 PM

P: 41

:D Maybe we have some misconception due to a different meaning of the word "indeterminate form". If you mean that [tex] \frac{\infty}{\infty} [/tex] is always an indeterminate form OK! It is an indeterminate form OK, but there are plenty of ways we can solve the indeterminateness. Also [tex] \frac{x^2}{x} [/tex] and [tex] \frac{x}{x^2} [/tex] and [tex] \frac{x}{x} [/tex] are the same type of indeterminate limits as they are [tex] \frac{\infty}{\infty} [/tex] but it all we know that they HAVE limit that are (respectively) [tex] \infty, 0, 1 [/tex] . [tex] \frac{ln(x)}{x} [/tex] tends to 0. I hate the De L'hopital tool but you can apply this and solve the matter yourself and see that it indeed goes to 0. Hope this helps. 


#23
Jul908, 05:20 PM

P: 41

Another proof.
It uses the elementry fact that [tex]e^y > y[/tex]. When y>0 you can apply the logaritm to that disequation and obtain [tex]y > \ln(y)[/tex] This hold for all [tex]y > 0[/tex]. Now when [tex]x \mapsto \infty[/tex] notice that [tex]0 < \frac{\ln(x)}{x} = \frac{2\ln(x)}{2x} =\frac{2\ln(\sqrt x)}{x} < \frac{2\sqrt x}{x} = \frac{2}{\sqrt x } \mapsto 0[/tex] 


#24
Jul908, 09:00 PM

P: 1,705

lol why did you bump this thread



#25
Jul1008, 02:54 PM

P: 41

Yeah! I just misunderstood everything. Now I read carefully there was no need to insist. The reason is I thought you were not sure that limit went to 0 so I used my argument, but in the end I understand you are perfecly aware and I just misunderstood the matter of the topic. I think I am not the only one who misunderstood. My apologize and see you next time! 


#26
Apr1610, 05:54 AM

P: 7

Hey guys I realise that this post is years late but I think its still useful for others to read ideas. I was looking at the start of the question which has
x^(1/x) I was wondering whether there was an easier way to prove that this tends to 1 as x tends to infinity. Can we not just focus on this part first: 1/x We can definitely show that this tends to 0 as x tends to infinity and so due to this no matter what x^(1/x) Will tend to 1 as its power tends to 0 


#27
Apr1610, 06:40 AM

HW Helper
P: 1,495

Your argument is false. Using your logic what would the limit be of (e^x)^(1/x) when x tends to infinity?



#28
Apr1610, 08:16 AM

P: 7

I would have thought that this would tend to e
Maybe I should have explained myself a bit better as my explanation was kinda just for the first example. Anything to the power of 1/x will tend to 1 as x tends to infinity so as in your example the whole answer does not tend to 1 but the part with x^(1/x) does and so the answer is e In a sense I am just looking at certain parts of the question before adding on any other part. Sorry if my explanation is confusing! Obviously if I'm wrong please tell me too. 


#29
Apr1610, 11:47 AM

HW Helper
P: 1,495

That explanation is incorrect as well, even though you find the correct answer. There is no x^(1/x) term in (e^x)^1/x. Remember (a^b)^c=a^(bc) and not a^(b^c).



#30
Apr1610, 12:06 PM

Emeritus
Sci Advisor
PF Gold
P: 16,092

We have a limit law that says [tex]\lim_{x \rightarrow a} f(x)^{g(x)} = \left( \lim_{x \rightarrow a} f(x) \right)^{\left(\lim_{x \rightarrow a} g(x) \right)}[/tex] whenever both limits on the r.h.s. exist, and the limits are in the domain of the exponentiation function, and exponentiation is continuous in a neighborhood of that point. [itex](+\infty)^0[/itex] isn't even in the domain of exponentiation! (If you don't know what an "extended real number is", then what I just said means that that's an indeterminate form) 


#31
Apr1610, 02:04 PM

P: 7

D'oh! Just wrote out loads of stuff only to realise I clicked on report instead of new post!
Again I realise that I haven't explained myself properly which is causing misunderstandings, completely my fault. To Cyosis: Thanks for pointing out to me that I am wrong in that part. I had completely forgotten that rule for some reason, silly me! Although in your question (e^x)^(1/x) this just equals e^(x/x),I was wondering however whether you could show me an example where a variable with only ^1/x does not tend to 1. I'm guessing it would be something with different variables used like n^(1/x) where n grows a lot faster than x? To Hurkyl: Sorry I wasn't trying to claim that (e^x)^(1/x) tends to 1, for some reason I thought what Cyosis pointed out to me that (a^b)^c=a^(bc) and not a^(b^c). P.S I don't want you guys to get the wrong impression of me, I'm not just trying to cause argument but get the grasp of all this so I can have a better understanding. Plus I think it's helping me to structure points better. 


#32
Apr1810, 06:57 PM

P: 20

To the people arguing the limit of ln(x)/x as x > infinity... that's just 0. Using l'hoptial's rule yields 1/x/1 = 1/x which goes to 0 as x goes to infinity...



#33
Apr1810, 11:10 PM

P: 1,105

Yeah dude you are about 3 years late.



#34
Apr1910, 02:30 AM

P: 308




#35
Apr1910, 01:07 PM

Emeritus
Sci Advisor
PF Gold
P: 16,092

Once you get the full statement of the theorem, please reference your favorite textbook or similar material for the definition of continuity and a statement about what domains upon which one might define an exponentiation operation, and where it might be continuous. 


#36
Apr1910, 03:58 PM

P: 308

Or do you have a "limit law" that it is not true?? 


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