shifting function with dense graph by identity function

Question

Let $f$ be a function on $\mathbb R.$ By $id$ I mean the identity function. Define $$(id\cdot f)(x)=id(x)\cdot f(x)=x\cdot f(x)$$ By graph of $f$, we mean $\Gamma(f)=\{\langle x,f(x)\rangle\colon x\in\mathbb{R} \}.$ If the graph of $\Gamma(f)$ is dense in $\mathbb{R}^2.$ Then the graph $\Gamma(id\cdot f)$ is dense in $\mathbb{R}^2.$ I think it true and I can see it geometrically since each rectangle in $\mathbb R^2$ will intersect the $\Gamma(f)$ and shifting by $\mathbb{R}$ will not change the density of $\Gamma(f)$. Is it correct or not ?

Answer 1

Consider any point $(a,b)$ with $a \neq 0$. Consider $(a,\frac b a)$. There exists a sequence $(x_n)$ such that $(x_n,f(x_n)) \to (a, \frac b a)$. This gives $x_n\to a$ and$f(x_n) \to \frac b a$. Hence $x_n f(x_n) \to b$ and $(x_n, x_nf(x_n)) \to (a,b)$. Can you finish?