# I A limit as t-->0 of log(t) / SQRT(t)

1. May 6, 2017

### Mr Davis 97

I am trying to find the following limit $\displaystyle \lim_{x \to 0^{+}} \frac{\log (t)}{\sqrt{t}}$, however, I don't see how. Obviously the answer is negative infinity, but I don't see how to get that. L'hospital's rule doesn't seem to work.

2. May 6, 2017

### andrewkirk

Use an epsilon-delta* argument, and take care with signs.

* Actually, where the limit is infinite, its an M-delta argument for big M.

3. May 6, 2017

### Orodruin

Staff Emeritus
Why would you use l'Hôpital? The numerator goes to minus infinity and the denominator to zero ...

4. May 6, 2017

### Mr Davis 97

Just to be clear, would that be the only way?

5. May 6, 2017

### Orodruin

Staff Emeritus
Ultimately, any way is based on the definition as proofs of all theorems regarding limits must be based on the definition of a limit. In your case thoguh, you have a product of a function whose limit is infinity and another whose limit is minus infinity.

6. May 6, 2017

### Mr Davis 97

Can we rigorously say that infinity times negative infinity is negative infinity?

7. May 6, 2017

### pwsnafu

Of course not, infinity is not a real number.

However you can rigorously prove: Let $f(x)$ diverge to infinity and $g(x)$ diverge to negative infinity as $x \to \infty$. Then $(fg)(x)$ diverges to negative infinity.
If you have done epsilon proofs, then you should be able to do this.

8. May 25, 2017

### phasacs

I don't see why not, you can write -∞ as a product of ∞ and -1 so you get -(∞2)=-∞

9. May 25, 2017

### Buffu

$\infty = \infty + 1 \iff \infty - \infty = 1 \iff 0 = 1$

Now ?

10. May 25, 2017

### Staff: Mentor

@Buffu, your example is unrelated to what @phasacs said, which by the way has an error.
If $\lim_{x \to a}f(x) = -\infty$ and $\lim_{x \to a} g(x) = -1$, then $\lim_{x \to a} f(x) g(x) = -\infty$. It wouldn't take much to prove this rigorously.

Last edited: May 25, 2017
11. May 25, 2017

### Staff: Mentor

It doesn't apply in this problem.
L'Hopital's Rule is applicable for limits of the form $\frac 0 0$ or the form $\frac{\pm \infty}\infty$.

12. May 25, 2017

### jack action

Wouldn't the proof be the following:
$$\lim_{t \to 0^{+}} \frac{\log (t)}{\sqrt{t}} = \lim_{t \to 0^{+}} \log (t)\frac{1}{\sqrt{t}}= \left( \lim_{t \to 0^{+}} \log (t)\right) \left(\lim_{t \to 0^{+}} \frac{1}{\sqrt{t}}\right) = (-\infty)(\infty) = -\infty$$

13. May 25, 2017

### Buffu

My example was to suggest that playing with infinity like a real number is not valid. Isn't that his error ?

14. May 25, 2017

### Buffu

I think somebody might argue with your proof as to how you know that "-ve infinity" times "+ve infinity" is "-ve infinity" ?

15. May 25, 2017

### Staff: Mentor

No, it isn't, although I don't know who you're referring to with "his error."
If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = \infty$, then it's fairly easy to rigorously show that
1. $\lim_{x \to a} f(x) \cdot (\pm g(x)) = \pm \infty$, and
2. $\lim_{x \to a} f(x) + g(x) = \infty$.
The limits you can't show are the ones that are indeterminate: $\lim_{x \to a} \frac{f(x)}{g(x)}$ and $\lim_{x \to a} f(x) - g(x)$.

16. May 25, 2017

### jack action

If I replace $\infty$ with a large number, then $-\infty \times \infty$ will get me a very large negative number (even if both $\infty$ are different values).

From what I know, infinity multiplication (with any real number or infinity) is not an indeterminate form. If you can find some reference stating otherwise, I would be glad to learn something new.

17. May 25, 2017

### Staff: Mentor

Right.

18. May 25, 2017

### Buffu

. You are correct.