Vtyjkyuk

I wanted to evaluate the limit
$$lim_xtoinftyx^2(1-cosfrac1x)$$

Since we know that
$-1leq cos xleq1$ and that $-1leq cosfrac1x leq 1$, so by algebraic manipulation,
$0leq x^2(1-cosfrac1x)leq2x^2 $.

Why does squeeze theorem fails to obtain a numerical limit here since the actual limit is $frac12$? I know the whole limit would be 0 when $x$ approaches 0.

Following up on my comment, one way to do it without Taylor expansions is so to note that
$$ cos 2y = 1-2sin^2 y quad Leftrightarrow quad 2sin^2 fracu2 = 1-cos u,$$
so
beginalign
lim_xto infty x^2left(1-cosfrac1xright) &=[u = 1/x] = lim_u to 0^+ frac1-cos uu^2\
&= lim_uto 0 ^+ frac2sin^2 fracu2u^2 = frac12lim_u to 0^+ fracsin^2 (u/2)(u/2)^2\
&= frac12 left(lim_u to 0^+ fracsin(u/2)u/2right)^2
= frac12cdot 1^2 = frac12.
endalign

This limit is fairly standard in high school: $;limlimits_uto 0dfrac1-cos uu^2=dfrac12$, so
$$x^2Bigl(1-cosfrac1xBigr)=frac1-coscfrac1xcfrac1x^2tofrac12 ;text as ;frac1xto 0.$$

$cos(t) = 1 - frac12 t^2 + O(t^4)$ for $t$ close to $0$ and, so when $|x|$ is large also for $cos(frac1x)$.

So substituting $t = frac1x$ we have that we consider the limit $frac1t^2(1 - cos(t))$ for $t downarrow 0$. Using the above estimate for $cos(t)$, the $1$'s cancel and we are left with $frac1t^2(frac12t^2 + O(t^4)) = frac12 + O(t^2)$ So indeed the limit is $frac12$ and series expansion is the easiest (IMHO) way to see it.