Solve the limit of fraction (tanx + sin2x)/ln(x+1)

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The required limit that x tends to is not specified. I take it to be 0 as substituting x = 0, makes the expression indefinite with 0/0.

We can use the l'Hopital's rule here and substitute the numerator and denominator with their derivatives.

=> lim x-->0 [ ((sec x)^2 + 2*cos 2x)/(1/(x+1))]

substituting x= 0

=> (1 + 2)/1

=> 3

The value of lim x--> 0[ (tan x + sin 2x)/ln(x+1)] = 3

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