/sci/ - Halp! — World4ch

Name: Anonymous 2008-09-22 18:12

I need help on a calculus problem. Does anyone know the answer (or can point me in the right direction) on this: What is the limit as x approaches 0 for tangent squared over x

Name: FFS 2008-09-22 18:13

Write the formula for the conjugate acid of the following bases:

CN-
HCO3-
NH3
PO43-

Name: Anonymous 2008-09-22 19:01

L'Hôpital's rule. lim(x->0) tan^2(x)/x = lim(x->0) 2sec^2(x) = 0? Check it with a graphing calculator, I could be wrong.

Name: Anonymous 2008-09-22 22:54

No, you're totally right. Thanks!

Name: Anonymous 2008-09-23 10:52

You can't even check it. Time paradox and such.

Name: Anonymous 2008-09-23 11:14

>>3

That's not quite right.

the derivative of tan^2(x) is 2sec^2(x)tan(x)

and sec(0) = 1.

so lim x-> 0 tan^2(x)/x = lim(x->0) 2sec^2(x)tan(x) = 0

So somehow, you managed to get the right answer, despite being pretty shit at maths.

How in the hell can you know L'Hopital's rule, and not know basic differentiation?

Name: Anonymous 2008-09-23 11:48

I actually realized that I forgot a tan(x), but didn't worry about reposting because the answer was still the same. Shitty differentiation...yea.

Name: Anonymous 2008-09-23 13:30

You all suck. Whenever you're doing a limit problem where x is going to zero and you've got trig functions, you can always replace sin(x) or tan(x) with x and evaluate the limit.

Tan(x)^2/x -> x^2 /x = x which goes to 0.

Name: Anonymous 2008-09-23 14:45

>>8

Yeah, good maths is just applying "rules of thumb" blindly.

You're an idiot.

Name: Anonymous 2008-09-23 15:04

Except that wasn't blindly applying anything. If you're too fucking stupid to show that the substitution is always valid then you should probably take calc1 again.

Name: Anonymous 2008-09-23 17:42

>>10

"I need help on a calculus problem. "

Probably best to teach this guy just to apply rules of thumb blindly?

I never took calc1, not an amerifag.

Name: Anonymous 2008-09-23 23:01

its .00000... 1 fucktard

Name: Anonymous 2008-09-24 0:36

it's easier to rewrite it as

= lim as x --> 0 (sinx/x * sinx/cos^2x)

= lim as x --> 0 (sin/x) * lim as x --> 0 (sinx/cos^2x)

= sinx/x goes to 1 times sinx/cos^2x goes to 0.

L'Hospital is fine too, just less work here I think.

Halp!

1 Name: Anonymous 2008-09-22 18:12

2 Name: FFS 2008-09-22 18:13

3 Name: Anonymous 2008-09-22 19:01

4 Name: Anonymous 2008-09-22 22:54

5 Name: Anonymous 2008-09-23 10:52

6 Name: Anonymous 2008-09-23 11:14

7 Name: Anonymous 2008-09-23 11:48

8 Name: Anonymous 2008-09-23 13:30

9 Name: Anonymous 2008-09-23 14:45

10 Name: Anonymous 2008-09-23 15:04

11 Name: Anonymous 2008-09-23 17:42

12 Name: Anonymous 2008-09-23 23:01

13 Name: Anonymous 2008-09-24 0:36