Difficult power limit

[wolly]

[math]\lim_{x \to \infty} x^{\left ( -\frac{1}{x} \right )}[/math]=

[math]\lim_{x \to \infty}\frac{1}{x^(\left{\frac{1}{x}} \right )}}[/math]=

[math]\lim_{x \to \infty}\frac{1}{e^{\ln x^(\left{\frac{1}{x}} \right)}}[/math]=

[math]\lim_{x \to \infty}\frac{1}{e^{\frac{\ln x}{x}}}[/math]
How can I prove that [math]\lim_{x \to \infty}\frac{lnx}{x}=0[/math]?
And If I substitute [math]x=e^{y}[/math] I get

[math]\lim_{y \to \infty}\frac{y}{e^{y}}[/math]?
I need to solve them without differentiation!

 

[wolly]

I Have no idea how to write in Latex the second and third limits.

 

[Dr.Peterson]

I Have no idea how to write in Latex the second and third limits.

You just had some wrong braces and parentheses, and used [ math ] instead of [ imath ], which made it hard to read.

Here is what you wrote, cleaned up:

[imath]\lim_{x \to \infty} x^{\left ( -\frac{1}{x} \right )}=\lim_{x \to \infty}\frac{1}{x^{\left(\frac{1}{x} \right )}}=\lim_{x \to \infty}\frac{1}{e^{\ln x^{\left(\frac{1}{x} \right)}}}=\lim_{x \to \infty}\frac{1}{e^{\frac{\ln x}{x}}}[/imath]​

​

How can I prove that [imath]\lim_{x \to \infty}\frac{lnx}{x}=0[/imath]?​

​

And If I substitute [imath]x=e^{y}[/imath] I get [imath]\lim_{y \to \infty}\frac{y}{e^{y}}[/imath]?​

​

I need to solve them without differentiation!​

I'll let someone else try to answer.

 

[wolly]

@Dr.Peterson You don't know how to solve them?
I wrote this in mathstackexchange and I got some answers

Difficult power limit $\lim_{x \to \infty} x^{\left ( -\frac{1}{x} \right )}$

This is my work so far $$\lim_{x \to \infty} x^{\left ( -\frac{1}{x} \right )}= \lim_{x \to \infty}\frac{1}{x^\left({\frac{1}{x}} \right)}=\lim_{x \to \infty}\frac{1}{e^{\ln x^\left({\frac{1}{x}}\r...

math.stackexchange.com

 

[lookagain]

@Dr.Peterson You don't know how to solve them?

Dr.Peterson did not necessarily claim that he did know how to solve them.
He deferred them to other forum users.

 

[mario99]

[imath]\displaystyle \lim_{x\rightarrow \infty}x^{-1/x} = 1[/imath]


[imath]\displaystyle \lim_{x\rightarrow \infty}x^{-1/x} = \lim_{x\rightarrow \infty}\frac{1}{x^{1/x}} = \lim_{x\rightarrow \infty}\frac{1}{e^{\frac{\ln x}{x}}} = 1[/imath]

This implies that the exponent [imath]\frac{\ln x}{x}[/imath] must approach [imath]0[/imath] as [imath]x \rightarrow \infty[/imath].

Then

[imath]\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x}{x} = 0[/imath]

 

[wolly]

@mario99 But isn't that [math]\frac{\infty}{\infty}[/math]?

 

[mario99]

@mario99 But isn't that [math]\frac{\infty}{\infty}[/math]?

[imath]\displaystyle \frac{\infty}{\infty}[/imath] could be anything. It was shown that it is zero in the case of that limit!

 

[khansaheb]

@mario99 But isn't that [math]\frac{\infty}{\infty}[/math]?

No - Apply L'Hospital's rule

 

[Dr.Peterson]

You don't know how to solve them?

I didn't have time at the moment to write and check a careful answer; I also wanted to separate my rewriting of your question from any answer that would be given. Just as in math, you shouldn't make assumptions ...

The central question in the OP was

How can I prove that [imath]\lim_{x \to \infty}\frac{\ln x}{x}=0[/imath]?
And If I substitute [imath]x=e^{y}[/imath] I get [imath]\lim_{y \to \infty}\frac{y}{e^{y}}[/imath]?
I need to solve them without differentiation!

The discussion has returned to that, but you ask

But isn't that [imath]\frac{\infty}{\infty}[/imath]?

It has that (indeterminate) form, but as you surely know, that just means you need to do something else to get a proper answer.

What you haven't told us is what facts you can use, since you don't want to use differentiation (e.g. L'Hopital's rule). It's been suggested that this limit is a basic one that students (who haven't yet learned to differentiate) might be given; but evidently you either haven't, or you were told that it is true, but want an actual proof (perhaps suspecting that the derivative might depend on this fact, which could be circular reasoning -- of course, it doesn't).

L'Hopital's rule is the natural thing to do, and is not "cheating"; it's what I would do. But it can be of interest to try to find limits without it -- when possible. On the other hand, it is pretty obvious from looking at the graph of [imath]\ln(x)[/imath] that it grows much slower than [imath]x[/imath], so your limit will be 0. That just isn't a proof.

So, again, what are you willing to assume for your proof? Some authors define the natural log as an integral (which sort of uses differentiation already); others define it as the inverse of the exponential function.

I'm not convinced by post #6; it seems to start with an answer to your initial problem, and then make a leap. But it raises the question whether you would be satisfied with a solution to [imath]\lim_{x \to \infty} x^{\left ( -\frac{1}{x} \right )}[/imath] that doesn't use [imath]\lim_{x \to \infty}\frac{\ln x}{x}[/imath].

Have you tried writing an epsilon-delta proof (well, epsilon-N proof, technically) of either limit?