Problem: find all $f$ such that $f(x^2 + f(y)) = f(xy)$ $f: \mathbb{R}\rightarrow \mathbb{R}$
My approach:
$\lim\limits_{n \to \infty} f((x/n)^2 + f(yn)) = f(xy) \rightarrow \textbf{f(x) = k}$
But we don't know whether the limit actually exists at $\infty$. Does anyone have any idea about what can be done?
(I'm assuming the function maps real numbers to real numbers.)
Let $k=f(0)$. Taking $y=0$ in the given equation tells us $$f(x^2+k)=k.$$ In particular, since the range of the function $x\mapsto x^2+k$ is the interval $[k,\infty)$, we have $$\tag{$\star$}f(z)=k\text{ for all }z\geq k.$$ Now, take any real $t$. Let $y\geq\max(k,0)$, and apply the given equation to $x=t/y$ to get $$f\left(\frac{t^2}{y^2}+f(y)\right)=f(t).$$ By $(\star)$, $f(y)=k$, and so, since $t^2/y^2\geq 0$, $t^2/y^2+f(y)\geq k$. This means that, by $(\star)$ again, the left side of the above equation equals $k$. So $f(t)=k$, as desired.
We can think of this solution as using the continuity of $x^2$ as opposed to the continuity of $f$, which isn't given. The substitution $y=0$ allows us to show that $f$ is constant on a long interval, and the rest is just using the equation somewhat naively to transport this property to all of $\mathbb R$.