Let $f(xy)=f(x)f(y)$ for all x and y. If the function is continuous at $x=1$, prove that $f(x)$ is continuous for $x$, for $x\neq 0$.
Since it is given $f(x)$ is continuous at 1 it implies $\lim\limits_{x \to 1}f(1-h)$=$\lim\limits_{x \to 1}f(1+h)$=$f(1)$ and also since $f(1)$=${f(1)^2}$ hence $f(1)$= 0 or 1
I got stuck and couldn't progress furthur
First, note that if $f(1) = 0$, then for all $x$, $f(x) = f(x) \times f(1) = 0$. So $f$ is zero everywhere, and therefore $f$ is continuous everywhere.
Else, $f(1) = 1$, but the idea is simple. Fix $x \neq 0$, and let $x_n$ be any sequence, $x_n \to x$.
We can assume that $x_n \neq 0$ for all $n$(because there can only be finitely many zeros in a sequence that converges to something non-zero, and removing finitely many terms of a sequence does not affect convergence).
Therefore , we obtain $\lim_{n \to \infty} \frac{x}{x_n} = 1$. Of course, this implies by continuity at $1$, that $\lim_{n \to \infty} f({x \over x_n}) = f(1)$. This implies that ${f(x) \over \lim_{n \to \infty} f(x_n)} = 1$, first by noting that $f(\frac {x}{x_n}) = \frac{f(x)}{f(x_n)}$ and then using the limit property for quotients.
The conclusion is clear from here, by the sequential definition of continuity.