efficiencie: Someone posted the below question, I quess Sir Chides and I think the doctors in the clinic need to diagnose and profer 'curative' measures for the problem:

Evaluate: ∫dx/(cosk + cosx)

doubleDx:

Yes, it's = 1

I think general efficiencie solved it correctly!...I will check out later. 1luv

I'm a little busy now!

STENON: My Quiz masters.....@Jackpot, Benbuks, Richiez......I dey greet all of una here oo

jackpot: Yea, the answer is 1, but I think the solution is correct to a ''great extent.''

jackpot: I am having problems with the bolded. How can you equate "limit" both sides and cancel out?

Next, that LimIny is as y tends to what?

STENON: My Quiz masters.....@Jackpot, Benbuks, Richiez......I dey greet all of una here oo

doubleDx:

Ok another approach. Rushed!

lim x→0 sin(x)^{sin(x)}

^^^

It's in indeterminate form of 0⁰.

Transforming yields => e^{[lim x→0 ln{sin(x)}.sin(x)]}

For indeterminate form of type 0.∞ =>

We write => lim x→0 ln{sin(x)}.sin(x) as => lim x→0 ln{sin(x)} ÷ 1/sin(x)
=>lim x→0 sin(x)ln{sin(x)}

Applying L'Hospital rule yields =>

=> e^{{lim x→0 -sin(x)}}
=> Factoring out constants yields =>

=> e^{- {lim x→0 sin(x)}}

=> e⁽⁰⁾

=> 1

benbuks:

yes dear , he's correct

though it ought to be

lny= -lim(sinx) & not " lim.lny =-lim(sinx) as written above .

=> lny =0 ..(as x-->0)

=> y=e^0

hence y=1

just realized your point , the guy try sha..

jackpot: I am having problems with the bolded. How can you equate "limit" both sides and cancel out?

Next, that LimIny is as y tends to what?

jackpot: it's a cheap question. K is a constant, so, cos k is also a constant. The rest follows similarly to the integration of sec x.

Hayormeah: how much glucose syrup with 20 % concentration has to be mixed with 100kg glucose syrup with 40 % concentration so that the mixture will have 36 % glucose?

doubleDx:

Ok another approach. Rushed!

lim x→0 sin(x)^{sin(x)}

^^^

It's in indeterminate form of 0⁰.

Transforming yields => e^{[lim x→0 ln{sin(x)}.sin(x)]}

For indeterminate form of type 0.∞ =>

We write => lim x→0 ln{sin(x)}.sin(x) as => lim x→0 ln{sin(x)} ÷ 1/sin(x)
=>lim x→0 sin(x)ln{sin(x)}

Applying L'Hospital rule yields =>

=> e^{{lim x→0 -sin(x)}}
=> Factoring out constants yields =>

=> e^{- {lim x→0 sin(x)}}

=> e⁽⁰⁾

=> 1

efficiencie:

if B=A(x) then the limiting value of A(x) as x tends to x(0) would also be the limiting value of B as x tends to x(0). Mathematically, limB=limA(x)=l as x→x(0) and hence B=A(x)=l in the limit. So i didn't cancel out the Lim's so to say, it simply follows from the definition of limits!


for limlny=-limsinx as x→0 the limiting variable is x and not y, since y is a function of x such that:
y=(sinx)^(sinx)

jackpot: now, from your bolded comment, if Lim_{x ->0}e^x = Lim_x->0cos x = 1, so can we say e^x= cos x?

jackpot:
outstanding.

But under which condition should one pass limit from ground to the exponent as you did in the bolded?

doubleDx:

Sorry, my bad; the error was from my typing it was supposed to be =>

lim x→0 sin(x)^{sin(x)}

^^^

It's in indeterminate form of 0⁰.

Transforming yields => lim x→0 e^{[ln{sin(x)}.sin(x)]}

For indeterminate form of type 0.∞ =>

We write => lim x→0 ln{sin(x)}.sin(x) as => lim x→0 ln{sin(x)} ÷ 1/sin(x)
=>lim x→0 sin(x)ln{sin(x)}

Applying L'Hospital rule yields =>

=> lim x→0 e^{-sin(x)}
=> Factoring out constants yields =>

=> lim x→0 e^{- {sin(x)}}

=> e⁽⁰⁾

=> 1


^^^^

Also if you use the continuity of e^sin(x) at x = 0, the bold can be rewritten as =>

1/lim x→0 e^{sin(x)} or e^{-{lim x→0 sin(x)}}.

The limit of sin(x) as x→0 is 0, hence the solution!

doubleDx:

Sorry, my bad; the error was from my typing it was supposed to be =>

lim x→0 sin(x)^{sin(x)}

^^^

It's in indeterminate form of 0⁰.

Transforming yields => lim x→0 e^{[ln{sin(x)}.sin(x)]}

For indeterminate form of type 0.∞ =>

We write => lim x→0 ln{sin(x)}.sin(x) as => lim x→0 ln{sin(x)} ÷ 1/sin(x)
=>lim x→0 sin(x)ln{sin(x)}

Applying L'Hospital rule yields =>

=> lim x→0 e^{-sin(x)}
=> Factoring out constants yields =>

=> lim x→0 e^{- {sin(x)}}

=> e⁽⁰⁾

=> 1


^^^^

Also if you use the continuity of e^sin(x) at x = 0, the bold can be rewritten as =>

1/lim x→0 e^{sin(x)} or e^{-{lim x→0 sin(x)}}.

The limit of sin(x) as x→0 is 0, hence the solution!

jackpot: now, from your bolded comment, if Lim_{x ->0}e^x = Lim_x->0cos x = 1, so can we say e^x= cos x?

jackpot: now, from your bolded comment, if Limx ->0e^x = Limx->0cos x = 1, so can we say e^x= cos x?

fasodecapo: I don't think so, since we can't say cos x is a function of e^x.

jackpot: exactly! Respect, Sire!

efficiencie:
cosx is clearly not a function of e^x, directly, but both are functions of 'x' and if both converge to unity as x tends to zero then in the limit they are equal such that cosx=e^x as x tends to zero.
hence if i say let y=e^x
then Limy=Lime^x as x tends to zero and if Lime^x=1 as x tends to zero then Limy=1 as x tends to zero and in other terms it could be stated that: e^x=1 as x tends to zero or y=1 as x tends to zero.
however with regards to my solution i guess i can make it clearer by adding these lines of reasoning:
Limlny=-Limsinx as x tends to zero
Limlny=0 as x tends to zero
lnLimy=0 as x tends to zero
Limy=e^0 as x tends to zero
Limy=1 as x tends to zero
and since y=(sinx)^(sinx)
then Lim[(sinx)^(sinx)]=1

jackpot: it's a cheap question. K is a constant, so, cos k is also a constant. The rest follows similarly to the integration of sec x.

jackpot: Dear Math Generals, Lieutenants and Sergeants,

Evaluate the limit: lim _{x -> 0} ^{(3x-tan 3x)}/_(x^3)

5Star tags: doubleDx, doubleDx, Laplacian, Richiez, benbuks, efficiencie, efficiencie, dejt4u, Amazing Angel, STENON, fasodecapo, Mathematical Scientists and Engineers.

jackpot: Dear Math Generals, Lieutenants and Sergeants,

Evaluate the limit: lim _{x -> 0} ^{(3x-tan 3x)}/_(x^3)

5Star tags: doubleDx, doubleDx, Laplacian, Richiez, benbuks, efficiencie, dejt4u, Amazing Angel, STENON, fasodecapo, Mathematical Scientists and Engineers.

efficiencie:


Let 'Lim' denote d limiting value as x tends to zero.

Lim{(3x-tan3x)/(x^3)}
On applyn l'hopitals, since the functn results in (0/0)
= Lim{(3-3.Sec^2(3x))/(3x^2)}
On applyn l'hopitals again, since the functn results in (0/0)
= Lim{(-18.Sec^2(3x)tan(3x))/(6x)}
(I luv dis!) On applyn l'hopitals again, since the functn results in (0/0)
= Lim{-9.sec^4(3x) - 18.Sec^2(3x)tan^2(3x)) }
= Lim{-9.sec^4(3x)} - Lim{18.Sec^2(3x)tan^2(3x)) }
= -9 - 18.(1).(0)
= -9

Auditors!

jackpot: that was exactly what I got, but. . .

doubleDx:

efficiencie got it correctly; it's -9. I will post mine later!