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 $0$.
 A: Rewrite the second term as $\frac{2}{y}\sum_{n\ge 2}{c_n}x^{2n}y^{-(n-2)}.$ Notice that the series itself is uniformly bounded in $y$ for all large enough $y$ (in fact, any $y>0$). Thus overall, you can upper bound bound the term by $\frac{C}{y}$ for some $C>0$, which goes to zero as $y\to\infty$.
A: $|\sum_{n=2}^\infty\frac{(-1)^n}{(2n!)}x^{2n}y^{-(n-1)}| \leq \sum_{n=2}^\infty\frac{1}{(2n!)}|x|^{2n} \frac 1 y$ for $y>1$. Note that $\sum_{n=2}^\infty\frac{1}{(2n!)}|x|^{2n}<\infty$.
[Look at the series expansion of $e^{|x|}$].
A: Since we can interchange limit and infinite sum when the sequence of functions converges uniformly we have that
$$\lim_{y\to\infty}2y(\cos(x/\sqrt y)-1)=\lim_{y\to\infty}(-x^2+2\sum_{n=2}^\infty\frac{(-1)^n}{(2n!)}x^{2n}y^{-(n-1)})=-x^2+2\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n!)}x^{2n}\lim_{y\to\infty}\frac{1}{y^{n+1}}=-x^2$$
A: $\cos(x) = 1 - \frac{x^2}{2} + O(x^4)$, 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.
A: $$\lim_{y \to \infty}2y\left(\cos \left(\frac{x}{\sqrt{y}}\right)-1\right)$$ Let $y=\frac 1{z^2}$ and the problem becomes
$$\lim_{z \to 0^+}\frac 2{z^2}\left(\cos \left({xz}\right)-1\right)$$ Let $x z=t$
$$\lim_{t \to 0} 2x^2 \frac{\cos (t)-1}{t^2}$$ Now, use equivalents or the very beginning of Taylor series.
A: If $x\ne 0$ then $$2y(-1+\cos (x/y^{1/2})=-4y \sin^2(x/2y^{1/2})=-4yz^2\cdot \frac {\sin^2 z}{z^2}$$ where $z= x/2y^{1/2}$. And $z\to 0$ as $y\to \infty.$
Footnote. From $\cos a=2\cos^2 (a/2)-1=1-2\sin^2(a/2)$ we obtain the half-angle formulas $\sin^2(a/2)=(1-\cos a)/2$ and $\cos^2(a/2)=(1+\cos a)/2.$
