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

[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.

 

[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.

 

[Orodruin]

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

 

[Mr Davis 97]

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

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

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

 

[Orodruin]

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

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.

 

[Mr Davis 97]

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.

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

 

[pwsnafu]

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

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.

 

[phasacs]

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

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

 

[Buffu]

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

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

Now ?

 

[Mark44]

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

@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.

 

[Mark44]

L'hospital's rule doesn't seem to work.

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

 

[jack action]

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

 

[Buffu]

@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.

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

 

[Buffu]

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

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

 

[Mark44]

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

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

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 $\lim_{x \to a} f(x) - g(x)$.

 

[jack action]

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

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.

 

[Mark44]

From what I know, infinity multiplication (with any real number or infinity) is not an indeterminate form.

Right.

 

[Buffu]

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.

. You are correct.