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Consider this proof of $\lim_{x\to{+\infty}} \frac{\ln(x)}{x} =0 $.

Using the simple substitution $x\mapsto x^n$ we have that $\forall n\in\mathbf{N}$ $$\lim_{x\to{+\infty}} \frac{\ln(x)}{x} = \lim_{x\to{+\infty}} \frac{n\cdot \ln(x)}{x^n}$$ and so we can write

$$\lim_{x\to{+\infty}} \frac{\ln(x)}{x} = \lim_{n\to{+\infty}} \lim_{x\to{+\infty}} \frac{n\cdot \ln(x)}{x^n}= \lim_{x\to{+\infty}} \lim_{n\to{+\infty}} \frac{n\cdot \ln(x)}{x^n}.$$

Now, the inner limit is 0 for all $x>1$ so it leaves $$\lim_{x\to{+\infty}} \frac{\ln(x)}{x^n} = \lim_{x\to{+\infty}} 0 = 0.$$

Is this proof correct?

user
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aleio1
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    A justification for why you can exchange limits might be harder than the proof of the original fact. – Maximilian Janisch Dec 06 '19 at 14:35
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    Also, the inner limit is essentially equivalent to the original limit, under the substitution $n=\ln(u)/\ln(x)$. – Barry Cipra Dec 06 '19 at 14:46
  • The easiest route to your limit is via the inequality $\log t\leq t-1, ,\forall t>0$. Putting $t=\sqrt{x} $ you get $\log x<2\sqrt{x}$. – Paramanand Singh Dec 06 '19 at 15:32
  • The meaning of the first equality of limits in your question is that "if one of the limits exists then so does other and they are equal". Your next step implicitly assumes that $\lim_{x\to\infty} (n\log x) /x^n$ exists and is independent of $n$. – Paramanand Singh Dec 06 '19 at 15:38

2 Answers2

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Personally, when two functions $g(x)$ and $f(x)$ both tend to infinity for $x→+∞$ but with different "speeds", I consider the hierarchy of infinites. The function $g(x)=x$ to the denominator dominates the function $f(x)=\ln x$ to the numerator because it goes faster than the $\ln x$ as you can see with this graph created with Desmos:

enter image description here

It's practically as if I had a constant at the numerator and an infinitely great function at the denominator. Therefore:

$$\lim_{x\to 0}\frac{\ln(x)}x=0$$

Sebastiano
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Your proof corresponds to use an exponential change of variable as $x=e^y$ with $y \to \infty$ that is

$$\lim_{x\to{+\infty}} \frac{\ln(x)}{x}=\lim_{y\to{+\infty}} \frac{y}{e^y}$$

and you are assuming that the latter, in the form $\frac n{x^n}$ in your case, is true but we also need to prove that fact.

To conlcude we can use that eventually $e^y\ge y^2$ which can be proved easily by induction for $y\in \mathbb N$ and then extended to reals.

user
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