1. ## Limits

Limits. Thank You.

2. Originally Posted by qbkr21
Limits. Thank You.
applying L'Hopital's a second time was not valid. you had the limit going to 1/0. for L'Hopital's we must have the limit going to $\frac 00$ or $\frac {\infty}{\infty}$

3. ## Re:

Suppose I have (#/0)...

Would the best solution be to take both the right and left hand limits of the # x is approaching...

Thks,

-qbkr21

4. Originally Posted by qbkr21
Suppose I have (#/0)...

Would the best solution be to take both the right and left hand limits of the # x is approaching...

Thks,

-qbkr21
Look at it this way:

we use the limit theorem, If $\lim f(x) = F$ and $\lim g(x) = G$, then $\lim f(x)g(x) = \lim f(x) \cdot \lim g(x) = FG$

thus, $\lim_{t \to 0} \frac {e^t}{3t^2} = \lim_{t \to 0} \frac {e^t}3 \cdot \lim_{t \to 0} \frac 1{t^2}$

and of course, you can get really formal with this, using sequential approaches to limits, but let's not go there, since they didn't ask us to "prove"

5. ## Re:

Re:

6. Originally Posted by qbkr21
Re:
well, $\frac {\infty}{0^{+}}$ makes no sense. but, no, $\frac {\infty}0$ and $\frac 0{\infty}$ are not indeterminate forms. they are "undefined" forms. there's a diference

7. ## Re:

Case in Point...

Thanks,

-qbkr21

8. Originally Posted by qbkr21
Case in Point...

Thanks,

-qbkr21
short of doing a proof for this question, i'd do something like this (i hope it's valid ):

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

since $\lim_{x \to 0^+} \frac 1x = \infty$ and $\lim_{x to 0^+} \ln x = - \infty$, that is, one is positive and the other is negative. we have that the products are negative and going to $- \infty$

thus $\lim_{x \to 0^+} \frac {\ln x}x = - \infty$

i know that's the answer, i don't know if what i did was 100% mathematically valid though, but it should be

TPH and Krizalid are the ones who are exceptional with limits

9. ## Re:

Cool kind of like setting up an irregular integral. I forgot that +INF * -INF = -INF

10. Originally Posted by qbkr21
Cool kind of like setting up an irregular integral. I forgot that +INF * -INF = -INF
yeah, that was the reasoning behind it

11. Originally Posted by Jhevon
short of doing a proof for this question, i'd do something like this (i hope it's valid ):

$\lim_{x \to 0^+} \frac {\ln x}x = \lim_{x \to 0^+} \frac 1x \cdot \lim_{x \to 0^+} \ln x$
I knew that we can split the limit if both of them by separate exist.

Originally Posted by Jhevon
TPH and Krizalid are the ones who are exceptional with limits
Well, TPH is better than me; and I'm a simple guy inloved of mathematics. I'm not in the college yet (that frustrates me)