# Limit of (1+ln x)/x as x -> 0

1. May 9, 2012

### Fre4k

1. The problem statement, all variables and given/known data

lim (1+ln x)/x = ?
x->0

2. Relevant equations

3. The attempt at a solution

lim 1/x + (ln x)/x
x->0
I know that (ln x)/x approaches -infinity faster than 1/x approaches infinity so the limit = -infinity, but how do I express this analytically?

2. May 9, 2012

### Fre4k

Ok, I got it. It's quite simple actually, duh.

3. May 9, 2012

### SammyS

Staff Emeritus
That's great to hear.

By the way, welcome to PF !

4. May 9, 2012

### sharks

Yep, you just had to use L'Hopital's Rule and the answer is infinity.

5. May 13, 2012

### Fre4k

Thanks. :)

6. May 13, 2012

### vela

Staff Emeritus
That's not an indeterminate form. The Hospital rule doesn't apply.

7. May 13, 2012

### sharks

Hi vela

Here is my understanding of the problem:
$$\lim_{n\to 0} \frac{1+\ln x}{x}=\lim_{n\to 0} \frac{1}{x}+\lim_{n\to 0}\frac{\ln x}{x}$$
$$\lim_{n\to 0} \frac{1}{x}=∞$$
$$\lim_{n\to 0}\frac{\ln x}{x}=\lim_{n\to 0}\frac{1}{x}=∞$$
The limit above is evaluated by using L'Hopital's Rule.

8. May 13, 2012

### vela

Staff Emeritus
That's not an indeterminate form. The Hospital rule doesn't apply.

9. May 13, 2012

### sharks

Agreed, -∞/0 is not an indeterminate form. I kept assuming n approaches ∞

To evaluate:
$$\lim_{n\to 0}\frac{\ln x}{x}=\lim_{n\to 0}\frac{1}{x}.\ln x$$
$$1/0=+∞$$
$$\ln 0=-∞$$
Hence, the product is -∞

Last edited: May 13, 2012
10. May 14, 2012

### skiller

So, so wrong.

11. May 15, 2012

### Infinitum

You wrote for n -> 0, but I assume that was a typo :tongue: Division by 0 is not defined for real numbers, so cannot exactly call it equal to positive infinity. Only,

$\lim_{x\to 0^{+}} \frac{1}{x} = +\infty$

$\lim_{x\to 0^{-}} \frac{1}{x} = -\infty$

Which means that the original limit will not be two sided.

12. May 15, 2012

### Steely Dan

The original limit is only meaningful approaching from the positive side, since the natural logarithm isn't defined on the negative side. So from the point of view of this problem, it is safe to say that the limit of $1 / x$ is positive infinity (unless we want to agree that a limit is only meaningful if it is defined from both sides, in which case this thread is over).

13. May 15, 2012

### sharks

Hi Infinitum and Steely Dan

Thank you both for the clarification.
$$\lim_{n\to 0^+} \frac{1+\ln x}{x}=\lim_{n\to 0^+} \frac{1}{x}+\lim_{n\to 0^+}\frac{1}{x}.\ln x=(+\infty)+(+\infty.-\infty)=(+\infty)+(-\infty)$$ which is undefined.

Last edited: May 15, 2012
14. May 15, 2012

### Steely Dan

This is not the way to go about it. If you know the properties of the natural logarithm, it's very clear that $\text{ln}(x) + 1 \approx \text{ln}(x)$ if $x$ is very small. So really this problem is just the limit of $\text{ln}(x) / x$ (which is clearly $-\infty$).

15. May 15, 2012

### sharks

I overlooked that part but it's clear that -∞+1≈-∞. Thanks again.