Determine all functions $f : \mathbb R \to \mathbb R$ satisfying the following two conditions:

(a) $f(x + y) + f(x - y) = 2f(x)f(y)$ for all $x, y \in \mathbb R$, and

(b) $\lim_{x\to \infty} f(x) = 0$.
1. Let \( P(x, y) \) denote the assertion \( f(x + y) + f(x - y) = 2f(x)f(y) \).
2. Let \( Q(x) \) denote the condition \( \lim_{x \to \infty} f(x) = 0 \).

### Case 1: \( f(0) = 0 \)
3. From \( P(0, 0) \), we have:
   \[
   f(0 + 0) + f(0 - 0) = 2f(0)f(0) \implies 2f(0) = 2f(0)^2 \implies f(0)(1 - f(0)) = 0
   \]
   Thus, \( f(0) = 0 \) or \( f(0) = 1 \).

4. Assume \( f(0) = 0 \). Then, from \( P(x, 0) \):
   \[
   f(x + 0) + f(x - 0) = 2f(x)f(0) \implies 2f(x) = 0 \implies f(x) = 0 \quad \forall x \in \mathbb{R}
   \]
   This is a valid solution. 

### Case 2: \( f(0) = 1 \)
5. Assume \( f(0) = 1 \). Then, from \( P(0, x) \):
   \[
   f(0 + x) + f(0 - x) = 2f(0)f(x) \implies f(x) + f(-x) = 2f(x) \implies f(-x) = f(x)
   \]
   This shows that \( f \) is an even function.

6. From \( P(x, -x) \):
   \[
   f(x - x) + f(x + x) = 2f(x)f(-x) \implies f(0) + f(2x) = 2f(x)^2 \implies 1 + f(2x) = 2f(x)^2
   \]

7. Using \( Q(2x) \), we have:
   \[
   \lim_{x \to \infty} f(2x) = 0
   \]
   Therefore:
   \[
   \lim_{x \to \infty} (1 + f(2x)) = 1 \quad \text{and} \quad \lim_{x \to \infty} 2f(x)^2 = 0
   \]
   This implies:
   \[
   1 = 0 \quad \text{(contradiction)}
   \]

Since the assumption \( f(0) = 1 \) leads to a contradiction, the only valid solution is \( f(0) = 0 \), which implies \( f(x) = 0 \) for all \( x \in \mathbb{R} \).

The final answer is \( \boxed{ f(x) = 0 } \) for all \( x \in \mathbb{R} \).