Evaluate the limit of the function ln(1+x)/(sinx+sin3x) x-->0

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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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)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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