# How to prove that $\ln(x)<x$ for $x\to \infty$?

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.

For the first part:

Note the following:

1. $\log(1) = 0 < 1$
2. $\log'(x) = \frac{1}{x} \leq 1 = \frac{d}{dx}x$ for $x \geq 1$
3. $\log'(x) = \frac{1}{x} \geq 1 = \frac{d}{dx}x$ 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 $n$.

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

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 $x > \ln(x)$, $\forall x > 1$.

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.

• I agree that this is a valid way to define $\ln x$, with many advantages. The alternative (as the inverse function of $e^x$) requires defining exactly what $e$ is, and this is no picnic. Stewart's popular calc book permits instructors to teach logs and exponentials in either order. I reluctantly taught exponentials first this fall so students would have more time to practice this material and score better on questions on it that appeared on the final exam (the FTC is towards the end of the semester). – Stefan Smith Dec 29 '13 at 2:09

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 $\ln$.

• you are right -- this is simpler than what I did -- can't think of everything I guess – Betty Mock Dec 29 '13 at 3:10
• @Betty Mock : mentioning Courant is some pretty impressive name-dropping! – Stefan Smith Dec 29 '13 at 3:56
• @Stephan Smith Didn't mean to drop names; I never actually met Courant; he was dead before I entered graduate school. However I had occasion awhile back to see what he wrote in his calculus book about $e^x$. It is so much easier to start with defining log x, and that provided an instant solution to a problem I had been hassling with for quite sometime. – Betty Mock Dec 30 '13 at 7:48