Continuity of the solution of this functional equation 
Let $f: \mathbb R \to \mathbb R$ be an additive function and $P:\mathbb R\to \mathbb R$ be a continuous function such that $$\forall x \neq 0, f(x)=P(x)f(1/x)$$
Prove that if $P(-1)>2$, $f$ is continuous.

$f$ being additive, it suffices to prove that $f$ is continuous at some point. I tried to proceed via contradiction, but it has led me nowhere.
I don't see how to use the $P(-1)>2$ hypothesis ...

Since $1/x$ is continuous in a neighbourhood of $-1$ with $1/(-1) = -1$, and $P(x)$ is continuous everywhere with $P(-1) > 2$, there is a $\delta > 0$, such that for all $x \in (-1 - \delta, -1 + \delta)$, $P(x) > 2$ and $P(1/x) > 2$. Hence, by the functional equation, for $x \in (-1 - \delta, -1 + \delta)$, you have
$$
|f(x)| \ge 2|f(1/x)| \ge 4|f(x)|.
$$
So $f(x)$ is identically zero on $(-1 - \delta, -1 + \delta)$ and hence is continuous at $-1$.
