How to prove that $\ln(x)During my calculus homework I need to prove some limits without using L'Hôpital's rule.
I have difficulties to show rigorously that  $\ln(x)<x$ for big enough x.
For example, I need to find the image of the continues function $f:R\to R$ such that for every $x\in R$: $|f(x)-x e^{\sqrt{\left| x\right| }}|<x^4$.
I've tried to prove that $xe^{\sqrt{\left| x\right|}}-x^4\to \infty$ if $x\to \infty$, and then the image of $f$ will be all the reals. Unfortunately, I don't find the way to make it formal enough. 
 A: Hint: Consider the function $f(x)=\ln(x)-x$. Note that $f(1)<0$ and show this function is monotonically decreasing (from $x=1$) by taking the derivative   
A: If we are allowed to take derivatives then we have:
$$\frac{\mathrm{d}}{\mathrm{d}x}(\ln(x))=\frac{1}{x} \text{ and } \frac{\mathrm{d}}{\mathrm{d}x}(x)=1$$
And we have that $\ln(1)=0 < 1$. Therefore we have $\ln(x)<x$, when $x=1$. If we examine the growth of the functions, we have that both are monotonically increasing with $x>1$ and as $\frac{1}{x}<1$, $x$ grows faster than $\ln(x)$ and therefore $x > \ln(x)$, $\forall x > 1$.
A: For the first part:
Note the following:


*

*$\log(1) = 0 < 1$

*$\log'(x) = \frac{1}{x} \leq 1 = \frac{d}{dx}x$ for $x \geq 1$

*$\log'(x) = \frac{1}{x} \geq 1 = \frac{d}{dx}x$ for $x \leq 1$.


So you can integrate over an appropriate interval to get $\log(x) < x$ for all $x$.
For the second part:
Try to use
$$
\lim_{x \to \infty} \frac{e^x}{x^n} = \infty
$$
for any natural number $n$.
A: Define ln x = $\int_1^x dt/t$  for x > 0. 
Then by the definition of the integral and the continuity of 1/x, ln x = (x-1)/c where  $1 \le c \le x$  if x > 1 and $1 \ge c \ge x$ if x < 1. The biggest (x-1)/c can be if x >1 is if c = 1, which gives x -1 < x.
If x < 1 the expresson (x-1)/c is negative; whereas between 0 and 1 x is positive, so log x < x in that case also.

Now you may think I'm playing games with this problem, but in fact that definition of ln x is given by Richard Courant in his calculus book.  Here is why he does it that way (besides that it makes problems like this one easy to do):
While usually the log is defined as the inverse of the exponential, that is a clumsy approach.  It leaves one with complicated and/or finicky proofs about the continuity and differentiability of the exponential.
However, if you define the log as ln x = $\int_1^x dt/t$ you gain many advantages.  All the properties of the logs can be derived directly and with little trouble from this definition. By this definition the log function is automatically differentiable.  Then when you define the exponential function as its inverse, all the properties you want for it fall out of the corresponding properties of the log, including, of course, its differentiability.
A: This is pretty similar to Betty's answer, but even simpler.  For $x \geq 1$, the comparison property of the definite integral gives
$$\ln x = \int_1^x \frac{1}{t}\,dt \leq \int_1^x 1\,dt = x-1<x.$$
The first equality is often used as a definition of $\ln$. 
