Find all functions with $f(x + y) + f(x - y) = 2 f(x) f(y)$ and $\lim\limits_{x\to\infty}f(x)=0$.

For any $x_0,y\in\mathbb{R}$, let $x=y+x_0$, then we have

$f(2y+x_0)+f(x_0)=2f(y+x_0)f(y)$

Hence,

$f(x_0) = 2f(y+x_0)f(y)-f(2y+x_0),\quad \forall y\in\mathbb{R}$

Consequently,

$\displaystyle f(x_0) = \lim_{y\to\infty}\left[f(x_0)\right] = \lim_{y\to\infty}\left[2f(y+x_0)f(y)-f(2y+x_0)\right] = 0$.

Using $x=y$, we have $f(2x) = 2f(x)^2 - 1$. Pick $X>0$ large enough so that $|f(x)| < 1/2$ for all $x \ge X$. Then $|f(2X)| > 1/2$, a contradiction.