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

[Fre4k]

Homework Statement

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

Homework Equations

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?

 

[Fre4k]

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

 

[SammyS]

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

That's great to hear.

By the way, welcome to PF !

 

[DryRun]

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

 

[Fre4k]

That's great to hear.

By the way, welcome to PF !

Thanks. :)

 

[vela]

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

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

 

[DryRun]

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.

 

[vela]

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

 

[DryRun]

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

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 -∞

 

[skiller]

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 -∞

So, so wrong.

 

[Infinitum]

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 -∞

You wrote for n -> 0, but I assume that was a typo 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.

 

[Steely Dan]

You wrote for n -> 0, but I assume that was a typo 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.

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).

 

[DryRun]

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.

 

[Steely Dan]

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.

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 $-\infty$).

 

[DryRun]

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 $-\infty$).

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