Find the limit please?
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"How do covalent bonds affect physical properties?"
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#
#lim_(x->0) (x^2 cot4x)/(sin2x) = 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#
