# Thread: [SOLVED] limit of the natural logarithm

1. ## [SOLVED] limit of the natural logarithm

How do you prove that $\lim_{x \to +\infty}ln(x) = +\infty$? Starting from the fact that $ln(x)=\int_1^x \frac{dt}{t}$, so without using anything related to $e^x$.
And also that $\lim_{x \to 0}ln(x) = -\infty$.
I just don't understand what my teacher did... Meaningless to me.

2. Originally Posted by arbolis
How do you prove that $\lim_{x \to +\infty}ln(x) = +\infty$? Starting from the fact that $ln(x)=\int_1^x \frac{dt}{t}$, so without using anything related to $e^x$.
And also that $\lim_{x \to 0}ln(x) = -\infty$.
I just don't understand what my teacher did... Meaningless to me.
Since $\frac{1}{t}$ is decreasing its min is at the right endpoint of any interval.

$
\int_{i}^{i+1}\frac{1}{t}dt > 1\cdot \left( \frac{1}{i+1} \right) \forall \\\ i \in \mathbb{N}
$

Summing both sides of the inequality gives

$\sum_{i=1}^{n}\int_{i}^{i+1}\frac{1}{t}dt > \sum_{i=1}^{n}\left( \frac{1}{i+1} \right)$

$\int_{1}^{n}\frac{1}{t}dt > \sum_{i=1}^{n}\frac{1}{i+1}=\infty$ as $n \to \infty$

3. Thanks! I understand what you did.

4. to see this

5. Thanks invite moon! My professor did something similar to this, now I understand him.
In your explanation, you should also show that $ln(x)$ is increasing, and that $ln3>1$.

6. Sorry ,I thought that ln(x) is increasing and that ln3>1,are well known things.

7. Yes there are. For instance, $ln(e)=1$, it can be shown that $e<3$. The derivative of $ln(x)$ is $\frac{1}{x}$, always positive in the dominion of the ln function, so ln is increasing.

8. Originally Posted by arbolis
How do you prove that $\lim_{x \to +\infty}ln(x) = +\infty$? Starting from the fact that $ln(x)=\int_1^x \frac{dt}{t}$, so without using anything related to $e^x$.
And also that $\lim_{x \to 0}ln(x) = -\infty$.
I just don't understand what my teacher did... Meaningless to me.
First note that natural logarithm is an incresing function (1) since $(\ln(x))'=\frac{1}{x}>0$ for all $x>0$

Now: $\ln(x)=\int_1^x\frac{du}{u}=\int_a^{a\cdot{x}}\fra c{du}{u}=\ln(a\cdot{x})-\ln(a)$

Thus the Natual Logarithm has the following property, for $a,b>0$: $\ln(a)+\ln(b)=\ln(a\cdot{b})$

Thus: $n\cdot{\ln(2)}=\underbrace{\ln(2)+...+\ln(2)}_n=\l n(2^n)$ for any natural number $n$ (2)

Suppose we want to prove that $\lim_{x \to +\infty}\ln(x) = +\infty
$
by definition.

We have to prove that, for all
$h>0$ there is a number $\delta_{h}>0$ such that if $x>\delta_{h}$ we have $\ln(x)>h$ (3)

Setting $\delta_{h}=2^{\frac{h}{\ln(2)}}$ by (1) and (2) we know that the condition (3) holds.

Just see that $\ln(2^n)=n\cdot{\ln(2)}>h$ iff $n>\frac{h}{\ln(2)}$ and since the function is increasing if $x>2^n$ then $\ln(x)>\ln(2^n)>h$

So by definition it follows that $\lim_{x \to +\infty}\ln(x) = +\infty
$
(4)

For the second part set
$u=\frac{1}{x}$ and use the property shown in the second line, so that $\ln(x)=-\ln(u)$ and since $x\to{0^+}$ we have $u \to +\infty$ and the assertion follows by (4)

9. Nicely done PaulRS. Thank you.

10. Originally Posted by TheEmptySet
$\sum_{i=1}^{n}\frac{1}{i+1}=\infty$ as $n \to \infty$
The problem with this approach is that the typical proof of the divergence of the harmonic series uses the integral test which uses $\lim_{x\to \infty}\ln x = \infty$. So it is kind of circular. But not really because there are some proofs which show divergence of harmonic series without integral tests.

Originally Posted by PaulRS
Setting $\delta_{h}=2^{\frac{h}{\ln(2)}}$ by (1) and (2) we know that the condition (3) holds.
Good job. But there is just a minor problem here. What does $2^{h/\ln 2}$ mean? You are assuming the definition of exponents for arbitrary real powers. However, the generalized definition needs further properties of $\ln$ which you are trying to avoid in the first place.

11. Originally Posted by arbolis
How do you prove that $\lim_{x \to +\infty}ln(x) = +\infty$? Starting from the fact that $ln(x)=\int_1^x \frac{dt}{t}$, so without using anything related to $e^x$.
i'll prove it using the definition of limit. let $M > 0.$ put $N=\lfloor{M \rfloor} + 1,$ and choose $K=2^{2N}.$ if $x > K,$ then:

$\ln x = \int_1^x \frac{dt}{t} > \int_1^K \frac{dt}{t} = \sum_{j=0}^{2N - 1} \int_{2^j} ^{2^{j+1}}\frac{dt}{t} > \sum_{j=0}^{2N-1} \int_{2^j}^{2^{j+1}} \frac{dt}{2^{j+1}}=\sum_{j=0}^{2N-1} \frac{1}{2}=N > M. \ \ \ \square$

And also that $\lim_{x \to 0}ln(x) = -\infty$.
I just don't understand what my teacher did... Meaningless to me.
let $\frac{1}{t}=u.$ then $\ln x = \int_1^x \frac{dt}{t} = -\int_1^{\frac{1}{x}} \frac{du}{u} = - \ln \left(\frac{1}{x}\right).$ now apply the first part of your problem.

12. Here is a general theorem. Let $f$ be a function (defined on $\mathbb{R}^+$) such that (i) $f$ is increasing (ii) $f$ is not bounded from above. Then $\lim_{x\to \infty}f(x) = \infty$. Here is the proof. We prove this directly by definition. Let $A>0$. Then since $f$ is not bounded from above there is $B>0$ such that $f(B)>A$. Then if $x>B$ it means $f(x) > f(B) > A$. Proof complete. Now we can apply this to $\ln$ function. Since $(\ln x)' = \tfrac{1}{x}>0$ this function is increasing this satisfies (i). Now we need to satisfy (ii) i.e. the function is unbounded this takes a little bit more work. Notice that (when $x>1$) $\smallint_1^x t^{-1}dt \geq \smallint_1^x t^{-3/2} dt = 2 -2x^{-1/2}$ and $2 - 2(9)^{-1/2} = \tfrac{4}{3} > 1$. Thus, $\ln (9) > 1$. Since $0= \ln 1<1$ and $\ln 9>1$ it means there is a number, call it $e$, such that $\ln e = 1$ (by property of continous functions). Then we see as PaulRS showed that $\ln e^n = n$ for all positive integers $n$. But since we can make $n$ are large as we please it means $\ln$ is an unbounded function. The rest follows.