GodOfWar_3ever said:
Hey people !
As each day passes, math keeps getting harder and harder -_-
First question -
lim x ----> 0 (e^x-1)/x
2nd question -
If 2A B = Π/4 (thats pi btw)
Prove that, tan B = (1 - 2 tan A - tan ^2 A)
1 plus tan A - tan ^2 A
Hence show that tan Π/8 is a root of the equation x^2 plus 2x - 1 = 0 and that its value is √2 - 1
If the other root is tan θ , find θ in the (0,Π) range.
Go !! (Need these with steps if possible)..
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First Q: the limit is 1, how you get there depends on the tools you're supposed to use
- if you know Taylor's expansion, then it's quite obvious as e^x=Sum_n(x^n/n!)=1 plus x plus x^2/2 plus x^3/6 ...= 1 plus x plus O(x^2). And thus: lim x->0 (x plus O(x^2))/x =1
- if you can't use Taylor, then there's L'Hopital rule, as both numerator and denominator tend to zero and are are differentiable in zero. Thus the limit is equal to the limit of their first derivatives, and it becomes lim x->0 of ((e^x-0)/1) =1
The second Q would requre me pen and paper, which I haven't atm... might update later.
Update: found a pencil, figured out you meant 2A plus B = pi/4, thus we go
- tan (2a plus b)= tan (pi/4)= 1
- tan (2a) plus tan(b) = 1- tan(2a)tan(b)
- 2tan(a)/(1-tan(a)^2) plus tan(b) = 1 - 2tan(a)tan(b)/(1-tan(a)^2)
- (1 plus 2tan(a)/(1-tan(a)^2))tan(b) = 1 - 2tan(a)/(1-tan(a)^2)
Divide by the factor of tan(b) and you got the desired expression.
Then, it's quite obvious than since tan(pi/4)=1 then
- tan(pi/8 plus pi/8) = 2tan(pi/8)/(1-tan(pi/8)^2) =1
Which gives your equation if you just call tan(pi/8)=x. Obviously pi/8 being in the first quadrant it has a positive tangent,so you have to choose the positive solution ie -1 plus sqrt(2) and not -1 -sqrt(2).
The negative solution is the opposite of the inverse of the first one, and is thus the tangent of your angle tranlated by pi/2, ie it's tan (5/8 pi). You could also reach the same result observing that tan(pi/4) = tan(pi/4 pi) = tan (5/4 pi) and thus the very same equation would be obtained for 5/8 pi. That's your theta.
