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Forums - General Discussion - Calculus HELP? Please

Vertigo-X on 29 August 2011
dany612 said:
Vertigo-X said:

delta x is simply the small difference between one point along a curve and the next. As in ∆x = x2-x1

 

I suppose I could take a stab at it.

 

lim ∆x->0 [(x+∆x)^2-x^2]/∆x

lim ∆x->0 [(x^2+2x∆x-∆x^2)-x^2]/∆x

lim ∆x->0 [(x^2-x^2)+2x∆x-∆x^2]/∆x

lim ∆x->0 [2x-∆x]∆x/∆x

lim ∆x->0 2x-∆x = 2x

 

f(x) = x^2

lim ∆x->0 [f(x + ∆x) - f(x)]/∆x


AKA the derivative!

y = nx^m

dy/dx = (n*m)x^(m-1)

 

The derivative is the easy way to do it. :)

Okay you lost me in the bolded. So what would be the limit? 


The bolded part is extraneous and unnecessary. I was just showing off. ;)

 

The answer is above the bolded.



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Vertigo-X on 29 August 2011
chapset said:
dany612 said:
pearljammer said:
ShadowSnake said:

Secondly to some of the responses being fed your way. If I remember correctly, you should not be telling dany about l'hopitals rule as that is a long ways from being taught to him.

Ah, you're right! I just checked and I hadn't been taught L'Hopital's rule until second semester of my first year.

so theres a more simplier way to solve it without the l'hospital rule?

L'Hopital is the simplest way, you do a derivative on top and a derivative in the bottom of your fraction, then you evaluate your limit

ex: lim x->0   (x^2 + 4x)/(x +4)

derivative on top = 2x + 4

derivative in the bottom = 1

then you evaluate your limit

lim x->0 (2x + 4)/1 = [2(0)+4]/1 = 4

 my english is not so great I hope you understand a bit better

Actually, I don't think L'Hopital's rule applies here. IIRC, it's a rule as a last resort; only use it if you can't simplify it beyond its current form.

 

lim x->0 (x^2 + 4x)/(x + 4)  can be simplified:

lim x->0 [x(x+4)]/(x+4)

lim x->0 (x) = 0



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Dark_Lord_2008 on 29 August 2011

Google it and you will find the answer.
lim x->0 sin(x)/x=1 lim x->0 1-cos(x)/x=0
http://answers.yahoo.com/question/index?qid=20080603183843AAHFO2l

lim->0 x+tan(x)/sin(x)

lim 1/ (x/sinx + tanx/sinx)

note:
lim sinx/x = 1
thus
lim x/sinx = 1

now, tanx / sinx = 1/cosx .
and lim (1/cos(x)) = 1, as x→0

thus
lim sinx/(x+ tanx) = 1/(1 + 1) = 1/2
(Is that the trig proof and answer?)








chapset on 29 August 2011
Vertigo-X said:
chapset said:
dany612 said:
pearljammer said:
ShadowSnake said:

Secondly to some of the responses being fed your way. If I remember correctly, you should not be telling dany about l'hopitals rule as that is a long ways from being taught to him.

Ah, you're right! I just checked and I hadn't been taught L'Hopital's rule until second semester of my first year.

so theres a more simplier way to solve it without the l'hospital rule?

L'Hopital is the simplest way, you do a derivative on top and a derivative in the bottom of your fraction, then you evaluate your limit

ex: lim x->0   (x^2 + 4x)/(x +4)

derivative on top = 2x + 4

derivative in the bottom = 1

then you evaluate your limit

lim x->0 (2x + 4)/1 = [2(0)+4]/1 = 4

 my english is not so great I hope you understand a bit better

Actually, I don't think L'Hopital's rule applies here. IIRC, it's a rule as a last resort; only use it if you can't simplify it beyond its current form.

 

lim x->0 (x^2 + 4x)/(x + 4)  can be simplified:

lim x->0 [x(x+4)]/(x+4)

lim x->0 (x) = 0

yes you are right I forgot about that



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Dark_Lord_2008 on 29 August 2011

What type of calculator does the OP need to solve these type of problems?
Input the question and the calculator spits out 8 to 10+ lines solving/proofing the trig proof.



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Vertigo-X on 29 August 2011
Dark_Lord_2008 said:

Google it and you will find the answer.
lim x->0 sin(x)/x=1 lim x->0 1-cos(x)/x=0
http://answers.yahoo.com/question/index?qid=20080603183843AAHFO2l

lim->0 x+tan(x)/sin(x)

lim 1/ (x/sinx + tanx/sinx)

note:
lim sinx/x = 1
thus
lim x/sinx = 1

now, tanx / sinx = 1/cosx .
and lim (1/cos(x)) = 1, as x→0

thus
lim sinx/(x+ tanx) = 1/(1 + 1) = 1/2
(Is that the trig proof and answer?)


That shouldn't work, either. I could be wrong, but this equation can be simplified further than it's original form:

 

lim x->0 x+tan(x)/sin(x)

lim x->0 x + [sin(x)/cos(x)]/sin(x)

lim x->0 x+ 1/cos(x)

Going straight up by your own rules, lim x->0 (0 + 1/1) = 1

 

EDIT: The thought occurred to me that I may have misinterpreted your equation. It should be written like this if x + tan(x) is the numerator:

lim x->0 [x+tan(x)]/sin(x)

 

EDIT2: In the case of that being your equation, I don't understand why you flipped the original limit in your work. The way I'd do it is:

 

lim x->0 [x+tan(x)]/sin(x)

lim x->0 [x/sin(x) + tan(x)/sin(x)]

lim x->0 x/sin(x) + lim x->0 tan(x)/sin(x)

(           1           ) + lim x->0 1/cos(x)

(1) + (1) = 2



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