13Is it =1, Manish sir ?
8 Answers
13y = (Sinx)tanx
ln(y) = SinxCosx. ln [1+ (sinx-1)]
Taking the limit & using series expansion fr ln(1+x)-
RHS= Lim(x→π/2) 1Cos(x)x [ (Sinx-1) - (Sinx-1)22 +....]
= Lim(x→π/2) (Sinx-1)Cosx + (Sinx-1)22.Cosx +....
After using L Hs'ptl,
Lim(x→π/2) ln(y) = Lim(x→π/2) -CosxSinx = 0
=> LIm(x→π/2) (y) = 1
1THIS IS 1 POWER INFINITY FORM. OR P POWER Q FORM.
THEN e power Q(P-1)
ans is i think so is 1
39Let x = π/2 - h where h is a very small positive number.
When x→π/2, h→0.
Therefore the limit changes accordingly as :
Limh→0(cos(h))cot(h)
= Limh→0[(1 + (cos(h) - 1))(cos(h)-1) ]cot(h)cos(h) - 1
= [ 1 ]Limh→0(cot(h)cos(h) - 1)
How can I avoid L'Hospital rule here? lol
I have no idea what I've done btw lola
13Yeah, i was thinking of the substitution too... x = π/2+h (h→0)
So, the expression-
ln(y) = tan(90+h). ln [1+ (cosh-1)]
= -cot(h). [ (cosh-1) - (cosh-1)22 +....]
=-1tan(h).[ (cosh-1) - (cosh-1)22 +....]
Finally using series Exp. again-
= -1(h + h3/3+..). [ (1-h2/2 +...-1) +...]
Which makes it come dwn to the form ≈ h/1 which again turns out 0 when (h-->0)
So, the Limit is =1.
1I can give a rather theoretical explanation . Continuing from Avik's post no. 3 , line 2 ,
ln y = tan x ( ln sinx )
Observe that " tan x " is a much slower increasing function than log ( sin x ) , which you can check by derivatives .
So , " tan x " approaches infinity much late than " log ( sin x ) " reaching zero ,when x tends to pi / 2 .
Hence , ln y = 0
Hence , y = 1
13Had myself used the same approach fr the answer, before getting dwn to pen it. ;)
Quick & confirming....gud u mentioned it Gallardo.