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

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2 Answers
Apr 2, 2018

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

Apr 2, 2018

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