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

**S1x** · 04-16-2007, 09:00 PM

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.

Chop Chop Master Onion · 04-16-2007, 09:05 PM

I'm drawing up a blank.

GURU ANT · 04-16-2007, 09:50 PM

Quote:

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** · 04-16-2007, 10:06 PM

***Ace Mercury*** · 04-16-2007, 10:35 PM

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*** · 04-17-2007, 02:00 AM

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!"

**Dr. Horrible** · 04-17-2007, 02:19 AM

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

***Ace Mercury*** · 04-17-2007, 07:57 AM

Quote:

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*** · 04-17-2007, 07:09 PM

^^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!"

**Dr. Horrible** · 04-18-2007, 11:12 AM

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*** · 04-23-2007, 08:37 AM

(courtesy of the interblag)

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

***CaptHayfever*** · 04-23-2007, 12:07 PM

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** · 04-23-2007, 01:33 PM

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

04-16-2007, 09:00 PM	#1
S1x VGF RPG Mod Join Date: Jul 2001 Location: Either over there or right here Gender: Posts: 5,057 Thanks: 15 Thanked 269 Times in 85 Posts Blog Entries: 1 Points: 136,453.61 Bank: 80,124.36 Total Points: 216,577.97 Donate	Don't often do this, but ARGH at this math problem 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.

04-23-2007, 08:37 AM	#11
*Ace Mercury* Marshmallow Knight ☆ Supermod Join Date: Jun 2000 Location: Southern Ontario Gender: Posts: 19,173 Thanks: 379 Thanked 1,657 Times in 837 Posts Blog Entries: 1 Points: 115,282.71 Bank: 31,567,564.03 Total Points: 31,682,846.74 Donate	Seems unnecessarily complicated. (courtesy of the interblag) Anyways, let us know what the answer is S1x, if you get it. There must be a trick.

04-16-2007, 09:05 PM	#2
Chop Chop Master Onion KICK, PUNCH, IT'S ALL IN YOUR MIND Join Date: Jun 2006 Location: Either UN's bed or Andre's bed. Gender: Posts: 13,965 Thanks: 982 Thanked 1,691 Times in 809 Posts Points: 48.31 Bank: 123,250.75 Total Points: 123,299.06 Donate	I'm drawing up a blank.

04-16-2007, 10:06 PM	#4
S1x VGF RPG Mod Join Date: Jul 2001 Location: Either over there or right here Gender: Posts: 5,057 Thanks: 15 Thanked 269 Times in 85 Posts Blog Entries: 1 Points: 136,453.61 Bank: 80,124.36 Total Points: 216,577.97 Donate

04-16-2007, 10:35 PM	#5
*Ace Mercury* Marshmallow Knight ☆ Supermod Join Date: Jun 2000 Location: Southern Ontario Gender: Posts: 19,173 Thanks: 379 Thanked 1,657 Times in 837 Posts Blog Entries: 1 Points: 115,282.71 Bank: 31,567,564.03 Total Points: 31,682,846.74 Donate	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.

04-17-2007, 02:00 AM	#6
*CaptHayfever* Smash Supermod Join Date: Jul 2002 Location: (n) - the place where I am Gender: Posts: 20,035 Thanks: 318 Thanked 1,018 Times in 616 Posts Points: 123,672.70 Bank: 2,638.67 Total Points: 126,311.36 Donate	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!"

04-17-2007, 02:19 AM	#7
Dr. Horrible PP&R Mod Current Events Mod Join Date: Jun 2000 Location: More important than where is when.... Gender: Posts: 6,787 Thanks: 137 Thanked 477 Times in 309 Posts Points: 6,637.82 Bank: 5,568,857.78 Total Points: 5,575,495.60 Donate	I'm pretty sure the derivative of Arctan(t) is 1/(t^2+1)

04-17-2007, 07:09 PM	#9
*CaptHayfever* Smash Supermod Join Date: Jul 2002 Location: (n) - the place where I am Gender: Posts: 20,035 Thanks: 318 Thanked 1,018 Times in 616 Posts Points: 123,672.70 Bank: 2,638.67 Total Points: 126,311.36 Donate	^^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!"

04-18-2007, 11:12 AM	#10
Dr. Horrible PP&R Mod Current Events Mod Join Date: Jun 2000 Location: More important than where is when.... Gender: Posts: 6,787 Thanks: 137 Thanked 477 Times in 309 Posts Points: 6,637.82 Bank: 5,568,857.78 Total Points: 5,575,495.60 Donate	For a minute I thought I had it with integration by parts and u substitution, but I just ended up proving 0=0.

04-23-2007, 12:07 PM	#12
*CaptHayfever* Smash Supermod Join Date: Jul 2002 Location: (n) - the place where I am Gender: Posts: 20,035 Thanks: 318 Thanked 1,018 Times in 616 Posts Points: 123,672.70 Bank: 2,638.67 Total Points: 126,311.36 Donate	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!"

04-23-2007, 01:33 PM	#13
S1x VGF RPG Mod Join Date: Jul 2001 Location: Either over there or right here Gender: Posts: 5,057 Thanks: 15 Thanked 269 Times in 85 Posts Blog Entries: 1 Points: 136,453.61 Bank: 80,124.36 Total Points: 216,577.97 Donate	I realized this a while ago; I just didn't say anything. Yeah, stupid easy question in disguise.