Proving the limit $\lim_{y \to \infty}2y(\cos(x/y^{1/2})-1)=-x^2$ I am wondering how one could prove the following limit:
$$\lim_{y \to \infty}2y(\cos(x/y^{1/2})-1)=-x^2$$
I tried using a series expansion:
$$\begin{aligned}2y(\cos(x/y^{1/2})-1)&=2y\sum_{n=1}^\infty\frac{(-1)^n}{(2n!)}x^{2n}y^{-n}=\\
&=2\sum_{n=1}^\infty\frac{(-1)^n}{(2n!)}x^{2n}y^{-(n-1)}=\\
&=-x^2+2\sum_{n=2}^\infty\frac{(-1)^n}{(2n!)}x^{2n}y^{-(n-1)}\end{aligned}$$
But I don't see how the second term, if it does, goes to $0$.

$\cos(x) = 1 - \frac{x^2}{2} + O(x^4)$, so $\cos\left(\frac{x}{y^{1/2}}\right) = 1 - \frac{x^2}{2y} + O(y^{-2})$.
Rearranging the terms gives
$2y\left(\cos\left(\frac{x}{y^{1/2}}\right) - 1\right) = -x^2 + O(y^{-1}).$ Taking the limit as $y \to \infty$ gives the answer.
