Question

How to find these type of question easily, without using L-hopitals rule?

Answer 1

`pi`

Explanation:

`sin(pi cos^2 x)/x^2 = sin(pi -pi sin^2)/x^2 = sin(pi sin^2 x)/x^2`

but for small `abs x` we have `sinx = x -x^3/(3!) + O(x^5)` and also

`sin^2 x = x^2+ O(x^4)` hence

`lim_(x->0)sin(pi cos^2 x)/x^2 =lim_(x->0)sin(pi(x^2+O(x^4)))/x^2 = pi lim_(x->0)sin(pi(x^2+O(x^4)))/(pix^2) = pi`

Answer 2

Please see below.

Explanation:

As Cesareo points out:

`sin(pi cos^2 x)/x^2 = sin(pi(1 - sin^2))/x^2`

`= sin(pi -pi sin^2)/x^2` `" "` (now use the difference formula)

`= sin(pi sin^2 x)/x^2`

`= sin(pi sin^2 x)/1 1/x^2`

`= sin(pi sin^2 x)/(pisin^2x) (pisin^2x)/x^2`

`= sin(theta)/(theta) pi (sinx/x)^2`

At `xrarr0`, we also have `theta rarr0`, so the limit is

`1(pi)(1)^2 = pi`