₦airaland Forum

ANDREW2EIC: I hail all the gurus in the house. I humbly plead you guys to help me solve these problems. 1.) if x = a(1-cosecD) and y = a(secD + tanD), show that xy² + a²(a² - x ) = 0.

2.) a polynomial p(x) is divided by x² - x and the remainder is A + Bx. Determine the constants Aand B. Detailed explanation would be appreciated.
Thanks in advance,

efficiencie: evaluate ∫(cosx/(2+cosx))dx

efficiencie: Lim(sinx/x) as x→∞

Put x=2πn+θ
Where
Limx as x→∞
is equivalent to
Lim(2πn+θ) as n→∞

Hence the new limit problem is:

Lim(sin(2πn+θ)/(2πn+θ)) as n→∞

Since sin(2πn+θ)=sinθ for all real values of 'n' no matter how large!

Then Limsin(2πn+θ)=Limsinθ as n→∞

= Lim(sin(θ)/(2πn+θ)) as n→∞

= sin(θ)/Lim(2πn+θ)) as n→∞

= sin(θ)/(Lim(2πn)+θ)) as n→∞

= sin(θ)/(Lim(2πn)+θ)) as n→∞

= 0

(I luv dis Jackpot sister, she's on fire!)

jackpot: Evaluate

lim_{x →∞}^{(sin x)}/_x

efficiencie: it should be in RADIANS sweet heart...the derivative and integral of trig functions are usually in radians nt degrees!

efficiencie: I tried it and f(x)→-9 as x→0 from the left and right but at x=0, f(0)→∞ where f(x)=(3x-tan3x)/x^3

efficiencie: lol! f(0)=0/0 na! abi!!

efficiencie: I don't have to check! The function
f(x) = (3x-tan3x)/x^3
records a 'jump discontinuity at x=0 because the left and right limits defined as:

Limf(x) as x→+0 and Limf(x) as x→-0 both converge to -9 but f(0)→∞ and it's expected that in the neighbourhood of 0, f(0)→0

efficiencie: but what, miss Jackpot!

efficiencie: but what, miss Jackpot!

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

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!

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

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 [color=deeppink]continuity[/color] 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 [color=deeppink]continuity[/color] 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!

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)

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

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

doubleDx: Yes, it's = 1

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

I'm a little busy now!

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)

efficiencie: Lim (sinx)^(sinx) as x→0
Let y=(sinx)^(sinx)
lny=sinx.lnsinx
lny=lnsinx/cosecx
Which results in -∞/∞ as x→0
Limlny=lim(lnsinx/cosecx)
Limlny=lim((dlnsinx/dx)/(dcosecx/dx))
Limlny=lim(cotx/-cosecxcotx)
Limlny=-lim(1/cosecx)
Limlny=-lim(sinx)
lny=0 as x→0
y=1
Hence
Lim (sinx)^(sinx)=1 as x→0

FrancisTony: *Sigh* He marked Economics not English, Mr. English professor!

Btw, some of the words you cancelled ain't wrong. You don't even know what you're doing. Smh!

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