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 =...

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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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