please solve limit class 12th
Answers
Step-by-step explanation:
y = lim x->0 (sinx/x)^(1/1-cosx)
=> lgy = limx->0 log(sinx/x) / (1-cosx)
=> logy = limx->0 (x/sinx)*(xcosx-sinx)/x² / (sinx)
=> logy = limx->0 (1/sinx/x) * (xcosx-sinx) /(x³sinx/x)
=> logy = limx->0 (1) * (xcosx-sinx)/x³*1
=> logy = limx->0 (cosx - xsinx - cosx) /3x²
=> logy = limx->0 -(xcosx + sinx) /6x
=> logy = limx->0 -(cosx - xsinx + cosx) /6
=> logy = -(1 - 0+1)/6
=> logy = -2/6
=> logy = -1/3
=> y = e^-1/3
Using the formula of limit we can say,
Now,
The limit is
Now,
Putting x=0 in this ,we get
The limit of this form is e^L
where
L=[f(x)-1].g(x) , where g(x) is the power
Putting x=0,in this we get
This is in indiscriminate form
Use L'hopital rule
Putting again x=0 we get 0/0,so diffrentiate it again
if we put 0 again,then we will again find 0/0 form
But this time we will try to apply formula
divide the numerator and denominator by x
we know,that
putting this value we get
now putting x=0,we get