Question

Find the limit please?

Answer 1

`= 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`

Answer 2

`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`