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 $0$.

Since we can interchange limit and infinite sum when the sequence of functions converges uniformly we have that
$$\lim_{y\to\infty}2y(\cos(x/\sqrt y)-1)=\lim_{y\to\infty}(-x^2+2\sum_{n=2}^\infty\frac{(-1)^n}{(2n!)}x^{2n}y^{-(n-1)})=-x^2+2\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n!)}x^{2n}\lim_{y\to\infty}\frac{1}{y^{n+1}}=-x^2$$
