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

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### 2 Answers

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**

You did not specified what is the accumulation point.

We'll suppose that x approaches to 0.

We'll solve the limit, creating remarcable limits.

lim tan x/x = 1, for x->0

lim sin x/x = 1, for x->0

lim ln(1+x)^(1/x) = ln e = 1

Now, we'll divide the numerator and denominator of the fraction by x:

lim (tanx + sin2x)/ln(x+1) = lim[ (tanx)/x + (sin2x)/x]/ln(x+1)/x

lim[ (tanx)/x + (sin2x)/x]/ln(x+1)/x = lim[(tanx)/x + lim(sin2x)/x]/lim ln(x+1)/x

We'll solve lim (sin2x)/x = lim 2*sin 2x/2x = 2lim sin 2x/2x = 2

lim (tanx + sin2x)/ln(x+1) = (1 + 2)/ln e = 3/1 = 3

**The limit of the given fractio, for x->0, is: lim (tanx + sin2x)/ln(x+1) = 3.**