# under the conditions given, prove $\lim\limits_{x\to 0} f(x)g(x)$ does exist.

$\lim\limits_{x\to 0} |f(x)|$ $\neq$ $\infty$ and if $\lim\limits_{x\to 0} f(x)$ exists, it's $= 0$.

Prove that there is a function $g$ such that $\lim\limits_{x\to 0} g(x)$ does not exist, but $\lim\limits_{x\to 0} f(x)g(x)$ does exist.

Or is this not true?

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The first thing that comes to mind is to let $g(x) = 1/f(x)$. The difficulty is points $x$ where $f(x)=0$. Let's try this: $$g(x) = \begin{cases} 1/f(x) & \text{if }f(x)\ne 0, \\ x & \text{if }f(x)=0. \end{cases}.$$ I thought maybe just $g(x)=1\text{ if }f(x)=0$, but the problem is that if $f(x)=0$ for all $x$ bigger than some number, then $\lim\limits_{x\to\infty}g(x)$ would exist.

If $a=\lim\limits_{x\to\infty}g(x)$ exists, then the set of points $x$ where $f(x)=0$ would have to be bounded, since otherwise $\limsup\limits_{x\to\infty} g(x)=\infty$, and it take "the limit exists" to mean that it's a real number, not $\infty$ or $-\infty$. So look at what happens after you get past all the points $x$ where $f(x)=0$, and you've got $\lim\limits_{x\to\infty} g(x)=a$, and hence $\lim\limits_{x\to\infty} f(x) = 1/a\ne0$, contradicting the hypothesis.

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The result is true.

Suppose first that $\lim_{x\to 0}f(x)=0$. Then let $g(x)=\sin(1/x)$ if $x\ne 0$. From here on we may assume that it is not the case that $\lim_{x\to 0}f(x)=0$.

Suppose next that there is a puntured neighbourhood $A$ of $0$ such that for every $a\in A$, we have $f(a)\ne 0$.

Let $g(x)$ be arbitrary outside $A$, and let $g(x)=1/f(x)$ for $x\in A$ and $x\ne 0$. Then $\lim_{x\to 0}f(x)g(x)=1$. But $\lim_{x\to 0}g(x)$ does not exist.

For if the limit is $0$, then $|f(x)|$ blows up as $x\to 0$, contradicting one of the stated conditions on $f(x)$. If the limit is $b$ where $b$ is non-zero real, then $\lim_{x\to 0}f(x)=1/b$, contradicting another of the conditions on $f$. And the limit cannot be some version of $\infty$, else we would have $\lim_{x\to 0}f(x)=0$, which has already been dealt with.

Finally, suppose that in every punctured neighbourhood of $0$, there is an $x\ne 0$ such that $f(x)=0$. Then let $g(x)=0$ where $f(x)\ne 0$, and $g(x)=\frac{1}{x}$ where $f(x)=0$.

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