Don't often do this, but ARGH at this math problem

[S1x]

I could draw an image, but I'm too lazy.

Find (the limit as x goes to 0) of:

1/x^3 * (integral from 0 to x) t^2 / t^4 + 1

Is this even solvable? I couldn't find the integral myself, and I tried entering it on something on the internet but got something crazy. Blargh. I already had to turn in the homework, but I just want to know wtf was going on here.

[Valigarmander]

I'm drawing up a blank.

[Eagle Eye]

Originally Posted by S1x

I could draw an image, but I'm too lazy.

Find (the limit as x goes to 0) of:

1/x^3 * (integral from 0 to x) t^2 / t^4 + 1

Is this even solvable? I couldn't find the integral myself, and I tried entering it on something on the internet but got something crazy. Blargh. I already had to turn in the homework, but I just want to know wtf was going on here.

Hmm.
Let me see, I'm terrible at limits, but I can do the integral for you.

I don't really understand what the problem is, so if this is wrong, let me know:

Use U substituion:
U = t^4 + 1
dU = 4t^3 + t

1/2t INT of dU/U

1/2t x ln(U)

Could be completely off but meh, can you explain that a bit better?

[S1x]

[Ace Mercury]

Once smug, I now feel powerless before integrals. I think there's some arctan stuff, if you want to evaluate the integral.

Final value theorem popped into my head, but I don't know if you're studying Laplace transforms yet. And... I can't do the Laplace transform of that.

[CaptHayfever]

That's a bugger of a problem, right there.
I'm going to assume you were given the dt at the end, elsewise the thing's completely invalid.

Um, I can't remember the exact form right now, and my calc book is downstairs, but have you tried integrating to arctan?

And remember, "I'm-a Luigi, number one!"

[Deku Trii]

I'm pretty sure the derivative of Arctan(t) is 1/(t^2+1)

[Ace Mercury]

Originally Posted by CaptHayfever

That's a bugger of a problem, right there.
I'm going to assume you were given the dt at the end, elsewise the thing's completely invalid.

Well, if it was dy, and t is a real number, it would evaluate to zero. Maybe it's a trick question

[CaptHayfever]

^^Actually, it would be dt/(t^2+1); don't forget your chain rule. But d(t^4) = 3(t^3), so I'm wrong anyway.

^Yeah, I'm with him.

And remember, "I'm-a Luigi, number one!"

[Deku Trii]

For a minute I thought I had it with integration by parts and u substitution, but I just ended up proving 0=0.

[Ace Mercury]

(courtesy of the interblag)

Anyways, let us know what the answer is S1x, if you get it. There must be a trick.

[CaptHayfever]

Thank you, Ace!

The limit does not exist. Whether it tends towards upper or lower infinity or not, I can't be certain at this point, though I would presume neither.

Why:
Finding limits this intricate, unless your teacher is a sadist who makes you do epsilon-delta proofs for each one, just involves evaluating at the value x approaches, in this case x->0. Having the cubed x in the denominator, though, is not a guaranteed infinite limit, as it might divide out or L'Hospital's rule might be applied. This is why I couldn't answer before solving the obscure integral form.
We can now see that the cubed x does NOT divide out, nor does the "numerator" evaluate to 0, so L'Hospital is out of the question. Our fate is sealed; there is no limit.
I'm inclined towards a full-blown DNE instead of +-infinity because of the logarithms. Put 0 into that first log term, and you're staring at log(-1), which is utterly preposterous, whether the base is 10 or e.

And remember, "I'm-a Luigi, number one!"

[S1x]

I realized this a while ago; I just didn't say anything. Yeah, stupid easy question in disguise.