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)