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.