Behaviour of $x^n$, $ln(x)$, and $e^x$ as $x\to \infty$

Question

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 $x$.

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

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.

Answer 1

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 $\ln x$.

Answer 2

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.

Answer 3

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?