Evaluate the limit of the function ln(1+x)/(sinx+sin3x) x-->0
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We have to find the value of lim x--> 0 [ ln(1+x)/(sinx+sin3x)]
substituting x = 0, we get the indeterminate form 0/0. Therefore we can use l'Hopital's Rule and substitute the numerator and denominator with their derivatives.
=> lim x--> 0 [ (1/(1+x))/(cos x + 3*cos 3x)]
Substitute x = 0
[ (1/(1+x))/(cos x + 3*cos 3x)]
=> (1 / 1) / ( 1 + 3)
=> 1/4
The required value of lim x--> 0 [ln(1+x)/(sinx+sin3x)] = (1/4)
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First, we'll substitute x by 0 into the function:
lim ln(1+x)/(sinx+sin3x) = ln 1/(sin 0 + sin 0) = 0/0
We've get an indetermination.
Since the trigonometric functions from denominator are matching, we'll transform the sum into a product:
sinx+sin3x = 2[sin (x+3x)/2][cos(x-3x)/2]
sinx+sin3x = 2sin 2x*cos(-x)
Since the cosine function is even, we'll put cos(-x) = cos x
sinx+sin3x = 2sin 2x*cos x
We'll re-write the limit:
lim ln(1+x)/(sinx+sin3x) = lim ln(1+x)/2sin 2x*cos x
We'll re-arrange the terms, crating remarcable limits:
(1/2)*lim ln(1+x)/sin 2x*lim (1/cos x)
But lim (1/cos x) = 1/cos 0 = 1
(1/2)*lim ln(1+x)/sin 2x= (1/2)lim [ln(1+x)/x]*(2x/sin 2x)*(x/2x)
(1/2)lim [ln(1+x)/x]*(2x/sin 2x)*(x/2x) = (1/2)*1*1*lim (x/2x)
(1/2)lim (x/2x) = 1/4
For x->0, lim ln(1+x)/(sinx+sin3x) = 1/4.
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