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In the chapter "Limits of a Function", I came across the following property:

As $x\to \infty$, $\ln(x)$ increases much slower than any positive power of $x$ where as $e^x$ increases much faster than any positive power of $x$.

So the following properties hold good:

$$(1) \lim_{x \to \infty} \frac{\ln(x)}{x}=0 $$

$$(2) \lim_{x \to \infty} \frac{(\ln(x))^n}{x}=0$$

$$(3)\lim_{x \to \infty} \frac{x}{e^x}=0$$

$$(4) \lim_{x \to \infty} \frac{x^n}{e^x}=0$$

For verifying the properties $(1)$ and $(3)$, I used the L'Hospital Rule, and I proved the limits tend to the value $0$.

I don't think the other two properties doesn't hold good at all conditions i.e., for all positive integral values of $n$. First of all, I was unable to use the L'Hospital Rule since I felt it would be very lengthy even if we know the value of $n$. So, I decided to use graphing calculator to determine their behaviour.

The following graph is for properties (1) and (2). The limit approaches $0$ at lower positive values of n but at a higher value say 98 as in the given graph. The limit itself approaches infinity and not zero. I tried to zoom out to see the behaviour, but as far as I tried the limit approaches infinity and not zero. Further from the graph it is evident the property given in my book is invalid, as the logarithmic function increases faster than the function $x$.

enter image description here

Similarly, I tried for the properties 3 and 4, as follows:

enter image description here

Clearly the property is again not working for higher values of $n$.

So at last, my doubt is:

Whether the property (Behaviour of $x^n$, $\ln(x)$, and $e^x$ as $x\to \infty$) given in my book correct for all values of $n$. If yes kindly verify or prove the properties 2 and 4. If no, kindly explain the reason.

Thanks in advance.

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4) follows by applying L'Hopital's Rule $n$ times. (You will end up with $\lim_{x\to \infty} \frac {n!} {e^{x}}$ which is $0$). 2) is same as 4) with $x$ changed to $\ln x$.

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  • $\begingroup$ Then the numerator of these limits eventually because $n!$. $\endgroup$ – jl00 Sep 19 at 6:02
  • $\begingroup$ Thank you for your answer sir. I understood. But, the graph of the function doesn't seem to tend to zero, instead tend to infinity. Why is this so, sir. Kindly explain. $\endgroup$ – M. Guru Vishnu Sep 19 at 6:05
  • $\begingroup$ @KaviRamaMurthy, I am not considering the case that $n$ approaching $\infty$. I noticed that, the property worked well for lower values of $n$ like $1,2,3,\dots$ but for larger constant values of n like $98$, the limit of the function as $x$ tends to infinity, tends to infinity and not zero. That is why I am confused. $\endgroup$ – M. Guru Vishnu Sep 19 at 6:16
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    $\begingroup$ @Intellex What your graphs show is that the convergence of $\frac {x^{n}} {e^{x}}$ to $0$ as $x \to \infty$ is not uniform w.r.t. $n$. $\endgroup$ – Kabo Murphy Sep 19 at 6:16
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    $\begingroup$ @Intellex Your graphs consider only small values of $x$ and they don't indicate what the limit as $x \to \infty$ is. The stated property is true for every positive integer $n$ (as long as $n$ does not depend on $x$). $\endgroup$ – Kabo Murphy Sep 19 at 6:19
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At first, as $n$ get greater, $e^x$ pass the $x^n$ later, so for graphing you must use appropriate domain. For $\ln x$ it's the same. But as @Kavi Rama Murthy said, you can use L'Hopital's Rule, as many as you need.

Also you can do it just once and use induction on $n$.

Notice it isn't the matter how big is $n$ or $n!$, as in each case they are finite.

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For number 4, recall

$$e^x=\sum_{i=0}^\infty \frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\dots+\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}+\dots$$

Consider $e^x-x^n$. Note the terms of degree $\geqslant n+1$ are unaffected by this difference. How will this affect your result?

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