Find the limit please?

We can use the first few terms of the various Taylor Series here:

#sin theta = theta - theta^3/(3!) + O(theta^5)#

Differentiate that to get the #cos# series:

#cos theta = 1 - theta^2/(2!) + O(theta^4)#

Replace #cot 4x# as #(cos 4x)/(sin 4x)#, set #theta = 2x# and #theta = 4x# as appropriate in the series expansions, and then insert the first few terms of those series to get:

#lim_(x to 0) (x^2 (1 - (4x)^2/(2!)))/(((4x) - (4x)^3/(3!))((2x) - (2x)^3/(3!))#

#= lim_(x to 0) (x^2 + O(x^4))/(8x^2 + O(x^4))#

#= lim_(x to 0) (1 + O(x^2))/(8 + O(x^2)) = 1/8#

Start by simplifying the expression considering that:

#cot4x = (cos4x)/(sin4x) = (cos^2 2x -sin^2 2x)/(2sin2xcos2x)#

so:

#(cot4x)/(sin2x) = (cos^2 2x -sin^2 2x)/(2sin^2 2xcos2x) = (cos2x)/(2sin^2 2 x) - 1/(2cos2x)#

Now we have:

#lim_(x->0) (x^2 cot4x)/(sin2x) = lim_(x->0) (x^2 cos2x)/(2sin^2 2 x) - x^2/(2cos2x)#

The second term is continuous in #x=0#:

#lim_(x->0) x^2/(2cos2x) = 0^2/2 = 0#

so:

#lim_(x->0) (x^2 cot4x)/(sin2x) = lim_(x->0) (x^2 cos2x)/(2sin^2 2 x) = lim_(x->0) ( cos2x)/(8((sin 2x)/(2x))^2#

and using the limit:

#lim_(theta->0) sintheta/theta = 1#

we can conclude:

#lim_(x->0) (x^2 cot4x)/(sin2x) = 1/8#