I have a question in which I want to show first that $\lim_{x\rightarrow\infty} \sqrt[x]{x}=1$ without L'Hôpital's rule/Integration/Series expantions or any other tool that isn't related to classic derivative material. We are given as a theorem that $\lim_{n\rightarrow\infty} \sqrt[n]{n}=1$ so I thought I could use Heine's definition. taking $x_n\rightarrow\infty$ I get

$$\left\lfloor x_{n}\right\rfloor ^{\frac{1}{\left\lfloor x_{n}\right\rfloor +1}}=\left\lfloor x_{n}\right\rfloor ^{\frac{1}{\left\lceil x_{n}\right\rceil }}\leq x_{n}^{\frac{1}{x_{n}}}<\left\lceil x_{n}\right\rceil ^{\frac{1}{\left\lfloor x_{n}\right\rfloor }}=\left(\left\lfloor x_{n}\right\rfloor +1\right)^{\frac{1}{\left\lfloor x_{n}\right\rfloor }}$$

I cannot continue from here though (to show the 2 sides converge to $1$, using $\lim_{n\rightarrow\infty} \sqrt[n]{n}=1$ ) , so I thank for all help in advance!

P.S This is to show that $\lim_{x\rightarrow\infty} \frac{\ln x}{x}=0$, so if anyone has a way for doing this one then it is welcome aswell.

  • $\begingroup$ You could point out that $\frac{d}{dx}\sqrt[x]x<0$ for $x>e$, so $\sqrt[x]x$ is monotonous for large $x$ and therefore follows the discrete limit. $\endgroup$ – Arthur Jan 20 '17 at 12:37
  • $\begingroup$ The "therefore" isn't straightfoward in my course at least as it is very formal. Moreover, the fact $f$ is decreasing (and positive) does not mean its limit is $1$, it just means it has a limit. $\endgroup$ – Theorem Jan 20 '17 at 12:42
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    $\begingroup$ The monotony means that $f(x)\in [f(\lceil x\rceil),f(\lfloor x\rfloor)]$, so one can use the known limit of $f(n)$ to say that $f(x)$ and $f(n)$ have the same limit. $\endgroup$ – Arthur Jan 20 '17 at 12:48
  • $\begingroup$ Sorry, I have not seen the word integration $\endgroup$ – Nosrati Jan 20 '17 at 13:09
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    $\begingroup$ If you know that $\sqrt[n]{n}\to1$ when the integer $n$ converges to infinity, use the fact that, for every real number $x>1$, there exists some positive integer $n_x\geqslant\frac12x$ such that $n_x\leqslant x\leqslant 2n_x$ hence $$\left(\sqrt[n_x]{n_x}\right)^{1/2}=\sqrt[2n_x]{n_x}\leqslant\sqrt[x]{x}\leqslant \sqrt[n_x]{2n_x}=\left(\sqrt[2n_x]{2n_x}\right)^2$$ and, since $n_x\geqslant\frac12x$, $n_x\to\infty$ hence $$\sqrt[n_x]{n_x}\to1\qquad\sqrt[2n_x]{2n_x}\to1$$ and you are done by the squeeze theorem since $1^{1/2}=1^2=1$. $\endgroup$ – Did Jan 20 '17 at 14:22

We are given as a theorem that $\lim\limits_{n\rightarrow\infty} \sqrt[n]{n}=1$ (...)

Excellent! In other words, we are given that $f(n,n)\to1$ when $n\to\infty$, $n$ being an integer, where, for every positive $(u,v)$, $$f(u,v)=v^{1/u}=\sqrt[u]{v}$$ We will use the fact (whose proof is left as an exercise) that, for every real number $x\geqslant1$, there exists some positive integer $n_x$ such that $$n_x\leqslant x\leqslant 2n_x$$ Then, $f(u,v)$ is increasing with respect to $v$ and decreasing with respect to $u$ for every $v>1$, hence $$f(n_x,n_x)^{1/2}=f(2n_x,n_x)\leqslant f(x,n_x)\leqslant f(x,x)\leqslant f(n_x,x)\leqslant f(n_x,2n_x)=f(2n_x,2n_x)^2$$ Since $n_x\geqslant\frac12x$, one knows that $n_x\to\infty$, and a fortiori $2n_x\to\infty$, which implies that $$\lim_{x\to\infty}f(n_x,n_x)=\lim_{x\to\infty}f(2n_x,2n_x)=1$$ hence we are done, by the squeeze theorem, since $1^{1/2}=1^2=1$.

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    $\begingroup$ Very nice, and as elementary as possible. I think the inequality $f(2n_x,n_x)\leq f(x,x)$ may not be obvious, but we have $f(2n_x, n_x)\leq f(x, n_x)\;$ (because $2n_x\geq x \geq 1$ and $n_x\geq 1),$ and $f(x,n_x)\leq f(x,x)$ because $1\leq x_n\leq x.$ $\endgroup$ – DanielWainfleet Jan 20 '17 at 19:42
  • $\begingroup$ @user254665 Indeed we have that, thanks for the comment. Added one intermediate inequality in each direction. $\endgroup$ – Did Jan 20 '17 at 19:48
  • $\begingroup$ Sometimes it's quite a challenge to derive a result without any advanced methods. $\endgroup$ – DanielWainfleet Jan 20 '17 at 19:54

The map $x\mapsto\frac{\ln(x)}{x}$ is decreasing on $[e,+\infty)$ and stays above $0$ on this interval.

So it has a finite limit $L$ as $x\to\infty$.

Using the fact that

$$\frac{\ln(x)}{x}=\frac{2}{\sqrt x}\,\frac{\ln(\sqrt x)}{\sqrt x}$$

and taking the limit as $x\to\infty$, we get :

$$L=0\times L$$

  • $\begingroup$ Okay I understood the solution however the conclusion you drew that it converges isn't a theorem. However it is for sequences. Is there an easy way to show that it has a limit? Perhaps Heine's definition? $\endgroup$ – Theorem Jan 20 '17 at 13:03
  • $\begingroup$ IMHO it is indeed a theorem, know as "monotonic limit theorem" (translation from french, sorry) $\endgroup$ – Adren Jan 20 '17 at 13:08
  • $\begingroup$ IMHO It should be a theorem in my course aswell but it isn't. We may only use theorems and questions solved in the books for these assignments so I have to prove this prior to using it as I wish. $\endgroup$ – Theorem Jan 20 '17 at 13:12

You mention that it is OK if one proves the following limit: $$\lim_{x \to \infty}\frac{\log x}{x} = 0\tag{1}$$ Needless to say before proving the above limit one must answer the following question:

What do you mean by the symbol $\log x$ when $x > 0$?

In other words it is essential to know the definition of $\log x$. Assuming that you know the definition of $\log x$ (which BTW is a non-trivial thing) you should be able to prove the following property of $\log x$: $$\log x \leq x - 1\tag{2}$$ for $x \geq 1$. Now $(2)$ can be used to prove $(1)$ via Squeeze Theorem.

Clearly if $x > 1$ then $\sqrt{x} > 1$ and hence by $(2)$ we have $$0 < \frac{1}{2}\log x = \log\sqrt{x} \leq \sqrt{x} - 1 < \sqrt{x}$$ or $$0 < \log x < 2\sqrt{x}$$ or $$0 < \frac{\log x}{x} < \frac{2}{\sqrt{x}}$$ and then using Squeeze theorem we get $$\lim_{x \to \infty}\frac{\log x}{x} = 0$$ On the other hand if my assumption (mentioned in bold earlier) is wrong then there is unfortunately no satisfactory way to even think about the limit $(1)$.


I will prove $\lim_{x\to\infty}\frac{\ln x}x = 0$ using only definition of limit and $e^x = \lim_n(1+\frac xn)^n$ (this is most likely the way you have it defined). Well, actually, to be fair, we do need basic properties of $\ln x$ and $e^x$, such that they are strictly increasing and inverse to each other.

Assuming $x>0$ we have:

$$\left(1+\frac xn\right)^n = 1 + \binom n 1\frac x n + \binom n 2\frac{x^2}{n^2}+\text{something positive}\geq 1+x+\frac{n-1}n\frac{x^2}2$$

and if we take limit on both sides, we get $$e^x\geq 1+ x+ \frac{x^2}2,\quad x>0.$$

We need to prove that

$$(\forall\varepsilon >0)(\exists M>0)\ x>M\implies \left|\frac{\ln x}x\right|<\varepsilon.$$

(Note: instead of $x>M$, I will just say "for $x$ large enough".)

Now, for large enough $x$,

$$\left|\frac{\ln x}x\right|<\varepsilon\iff\ln x<\varepsilon x\iff e^{\varepsilon x}>x$$

and since $e^{\varepsilon x}\geq 1 + \varepsilon x + \frac{(\varepsilon x)^2}2$, it is enough to show that $$1 + \varepsilon x + \frac{(\varepsilon x)^2}2>x$$ for large enough $x$. This is obviously true since we have "upward" parabola, but to be thorough, you could divide by $x$ and use axiom of Archimedes to show that $$\frac{\varepsilon^2}2x>1$$ for large enough $x$.


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