Functional equations [Sample paper of Indian Mathematical Olympiad] Edit- There was information missing (lack of clear printing in my book) in the book through which I referred the question. Confirming with my friend's book I have made a small change. I am really sorry.
Edit 2- Guys, this question is meant for an olympiad where calculus isn't allowed. Try solving it without calculus.

If $f: \mathbb{R} \to \mathbb{R}^+$ satisfies:

$f(xf(y)) = yf(x)$ for all $x,y \in\Bbb{R}^+$
$\lim_{x \to \infty} f(x) = 0$

Find $f(x)$.

I tried putting $x=y$ and got $f(y(f(y))=yf(y)$ which indicates $f(x) = x$ which doesn't follow the second statement. Help.
Sadly I don't have enough rep to comment:

With the question as written do we not get $f(x) = 0$ for all $x$? It seems inputting any negative value of $y$ and using that $\text{Im} f \subset [0,\infty)$ leads to this conclusion.
Assume the domain $f$ is $\mathbb{R}^+ = \{x \in \mathbb{R}: x &gt; 0\}$, not $\mathbb{R}$. This approach is similar to Mishikumo2019's.

In the functional equation $f(x f(y)) = y f(x)$, setting $x = 1$ and rewriting $y$ as $x$ gives $$f(f(x)) = x f(1).$$ The RHS increases without bound but $f(x) \to 0$ as $x \to \infty$, so for every value $Y \in \mathbb{R}^+$ there exists some $\delta &gt; 0$ such that if $x &lt; \delta$ and $x \in f(\mathbb{R}^+)$, then $f(x) &gt; Y$: that is, $\lim_{x \to 0^+} f(x) = +\infty$ if the domain of $f$ is restricted to the range of $f$. (With the additional hypothesis that $f$ is continuous, this is equivalent to $\lim_{x \to 0^+} f(x) = +\infty$.)

We know from setting $x = y$ in the functional equation that any number of the form $x f(x)$ is a fixed point of $f$; as a corollary, $f$ has at least one fixed point. If $x_0$ is a fixed point, then so is $x_0 f(x_0) = x_0^2$ and, more generally, $x_0^{2^n}$ for any $n \in \mathbb{N}$. If $x_0 \neq 1$, then we have a sequence of fixed points approaching zero (contradicting the fact that $\lim_{x \to 0^+} f(x) = +\infty$ when the domain of $f$ is restricted to the range of $f$) or extending out to infinity (contradicting $\lim_{x \to \infty} f(x) = 0$). Thus, the only fixed point of $f$ is $1$, so $x f(x) = 1$ for all $x \in \mathbb{R}^+$, and thus the only possible $f$ is $f(x) = 1/x$.