Prove $\lim_{x\rightarrow\infty} \sqrt[x]{x}=1$ (No L'Hôpital)

Question

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.

Answer 1

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 $1^{1/2}=1^2=1$.

Answer 2

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 :

$$L=0\times L$$

Answer 3

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 $x$.

Answer 4

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 $(1)$.