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I want to find

$$\lim_{x \to \infty} \frac{x}{\log(x)}$$

I can see that it will be infinity as $x$ increases too fast then logx. But how can I find it mathematically? I tried using lhospital rule but as definition of lhospital rule says that it is allowed to use if answer of limit exists, but here answer is infinity so I think that I can not use it. Please correct me in case I am wrong. Thank you.

Edit1:

Can we use lhospita here or not. Will using lhospital will be valid or not.

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  • $\begingroup$ yes l'Hospital is valid! $\endgroup$ – gimusi Dec 16 '17 at 8:19
  • $\begingroup$ Since you are facing with an indeterminate form $\frac{\infty}{\infty}$ $\endgroup$ – gimusi Dec 16 '17 at 8:21
  • $\begingroup$ But definition of lhospital rule says that we can use it only if result of lim exist. And in this case infinity does not exist. $\endgroup$ – golushailuoo7 Dec 16 '17 at 9:06
  • $\begingroup$ @goloshailuoo It must be interpreted in a general sense that is that limit exist finite or infinite. Take also a look here math.stackexchange.com/questions/990941/… $\endgroup$ – gimusi Dec 16 '17 at 9:13
  • $\begingroup$ Limits equals to $\pm \infty$ are completely free to exist. When we think to a not existing limit we are dealing with limit such $sinx$ as $x+\infty$. In this case you can't define any limit. $\endgroup$ – gimusi Dec 16 '17 at 9:18
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$$x=\exp(\log x)=1+\log x +\frac{(\log x)^2}2+\cdots>\frac{(\log x)^2}2$$ if $x>1$ and so $$\frac x{\log x}>\frac{\log x}2$$ etc.

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  • $\begingroup$ Thanks for the solution, but can I use lhospital rule or not? $\endgroup$ – golushailuoo7 Dec 16 '17 at 8:08
  • $\begingroup$ @golushailuoo7 yes you can apply l'Hospital! $\endgroup$ – gimusi Dec 16 '17 at 9:35
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Suppose $x\gt 4$ then $$\frac {x^2}{\log {x^2}}=\frac x2\cdot\frac x{\log x}\gt2\cdot \frac x{\log x}$$You can use this to show that the function is unbounded as $x\to \infty$

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By substitution

$$\frac x{\log x}=\frac {e^{\log x}}{\log x}=$$

set $\log x = y \to +\infty$

$$=\frac {e^{y}}{y}\stackrel{\text{definitively}}>\frac {y^2}{y}=y\to+\infty$$

You can also use l'Hospital:

$$\frac x{\log x} \stackrel{\text{Hospital}}\implies\frac 1{1/x}=x\to+\infty$$

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  • $\begingroup$ Gimusi. You are #1 !:)))P.S. I swear I did not copy!!!! $\endgroup$ – Peter Szilas Dec 16 '17 at 9:42
  • $\begingroup$ In any case we can say: "Repetita Juvant"!!! ;) $\endgroup$ – gimusi Dec 16 '17 at 9:44
  • $\begingroup$ Gimusi. Bravo!!! $\endgroup$ – Peter Szilas Dec 16 '17 at 9:52
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$x \rightarrow +\infty.$

Let $x:=e^y; $

exponential fct. is bijective with $D= \mathbb{R}$, $R= (0,\infty).$

Consider for $y \gt 0$:

$F(y): =\dfrac{e^y}{y} = \dfrac{1+y + y^2/2! +...}{y}$

$\gt y/2.$

$\lim_{y \rightarrow \infty} F(y) =\infty.$

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  • $\begingroup$ Nice Peter, that's also my solution without l'Hospital! $\endgroup$ – gimusi Dec 16 '17 at 9:37
  • $\begingroup$ Thanks, gimusi. I noticed the similarity. Hopefully presented a bit differently. $\endgroup$ – Peter Szilas Dec 16 '17 at 9:40
  • $\begingroup$ Yes you have motivated in more detail why $e^y>y$ definitively. $\endgroup$ – gimusi Dec 16 '17 at 9:43
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    $\begingroup$ That was the easy part.:)) $\endgroup$ – Peter Szilas Dec 16 '17 at 9:56
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May I suggest this is a simple high school limit?

One learns in high school that $$\lim_{x\to \infty}\frac{\log x}x=0^+$$ so the answer is obvious by the general rules for limits.

This standard limit is obtained proving that $$\forall x>1,\;0<\log x<2\sqrt x\quad\text{whence}\quad 0<\frac{\log x}x <\frac1{\sqrt x}.$$

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