# What is the limit of the fraction tan4x/tan2x, x-->0 ?

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We need to find the value of lim x-->0 [ tan 4x / tan 2x]

If we substitute x = 0, we get the indeterminate form 0/0. This allows us to use l'Hopital's rule and substitute the numerator and the denominator with their derivatives.

=> lim x-->0 [ 4*(sec 4x)^2 / 2* (sec 2x)^2]

substitute x = 0

=> 4*1 / 2*1

=> 4/2

=> 2

**The required value of lim x-->0 [ tan 4x / tan 2x] = 2.**

We'll create the remarcable limits:

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

We'll re-write the function:

lim [4x*(tan 4x)/4x]*[(2x)/2x*tan 2x] = lim 4x*lim [(tan 4x)/4x]*lim[(2x)/tan 2x]*lim (1/2x)

We know that lim [(tan 4x)/4x] = 1 and lim[(2x)/tan 2x] = 1

lim [4x*(tan 4x)/4x]*[(2x)/2x*tan 2x] = lim 4x*lim (1/2x)

lim [4x*(tan 4x)/4x]*[(2x)/2x*tan 2x] = (4/2)lim(x/x)

**The limit of the given function is : lim tan4x/tan2x = 2, if x -> 0.**