Question

Lim ((sin(pi(cosx)))/(x^2)) as x->0

Answer 1

Both the numerator and denominator approach 0, so by L'Hopital's rule

`\displaystyle\lim_(x\to0)(\sin(\pi\cos x))/(x^2)=\lim_(x\to0)(-\pi\sin x\cos(\pi\cos x))/(2x)`

Using the fact that

`\displaystyle\lim_(x\to0)\frac{\sin x}x=1`

we find

`\displaystyle\lim_(x\to0)(-\pi\sin x\cos(\pi\cos x))/(2x)=\lim_(x\to0)\frac{-\pi\cos(\pi\cos x)}2`

`\pi\cos x\to\pi` as
`x\to0`, and
`\cos\pi=-1`, so the limit is π/2.

(I'm wondering if there's a way to do this without L'Hopital's rule, in case you haven't learned it yet...)