# Limit problem function

Prove that the statement is wrong.

If $\lim f(x)$ exists but $\lim g(x)$ does not exist, then $\lim f(x)g(x)$ does not exist either.

Consider $f(x) = \dfrac1x$, $g(x) = (-1)^{\lfloor x \rfloor}$. Consider $x \to \infty$ of $f(x)$, $g(x)$ and $f(x)g(x)$.
• And if Andi would like $g$ to be continuous, $g(x)=\sin x$ does nicely. – Brian M. Scott Dec 16 '13 at 22:11
Consider $f(x) = 0$ and $g(x) = 1/x$, and look at the limits as $x \rightarrow 0$.
Consider the functions $$f(x) = 0 \quad \quad g(x) = x$$ and their infinite limits. Clearly $$\lim_{x \rightarrow \infty} f(x) = 0$$ Note that $\lim_{x \rightarrow \infty} g(x)$ does not exist. However, note $$f(x)g(x) = 0$$ Thus, $\lim_{n \rightarrow \infty} f(x)g(x) = 0$, which exists.
Less trivial, as $\ x\to \infty,\$ $\ 1/x \to 0\,$ and $\,\sin(x)$ doesn't exist, but $\ \sin(x)/x\to 0$