Find the limit please?

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2 Answers
Mar 18, 2017

= 1/8

Explanation:

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

Mar 18, 2017

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

Explanation:

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