Proving $\lim_{x\to x_{0}} f(x) = 0$ implies $\lim_{x\to x_{0}} f(x)g(x) = 0$ in $\mathbb{R}^{n}$ Let $A$ be a subset of $\mathbb{R}^{n}$ and let $x_{\star} \in \mathbb{R}^{n}$ Also, suppose $g : A \rightarrow \mathbb{R}$ is bounded; that is, there is a real number $c$ so that $|g(x)| \leq c$ for all $x \in A$. Prove $\lim_{x\to x_{\star}} f(x) = 0$, then $\lim_{x\to x_{\star}} [g(x)f(x)] = 0$.

Here's my try:
Since $|g(x)| \leq c$ for all $x$, we have $|g(x)f(x)| \leq c|f(x)|$. Therefore, $\lim_{x\to x_{\star}} g(x)f(x) \leq \lim_{x\to x_{0}} c\cdot f(x) = 0$, so we are done.
For some reason, I think this is wrong. I think the formal way to do it is to introduce a sequence with limit point $x_{\star}$, but I have no clue how to do it this way. Can someone please help me?
 A: You have the right idea, but your argument is technically incorrect. You write at one point $\lim_{x\to x_\star}g(x)f(x)\le \lim_{x\to x_{\star}}c\cdot f(x)$, but we have not yet justified that the limit $\lim_{x\to x_\star}g(x)f(x)$ exists, so it is not correct to put it in an inequality.
A formal way to prove it is to note, as you did, that $|g(x)|\le c$ for all $x\in A$, and let $\epsilon > 0$ be a challenge number. Then since $f(x)\to 0$, there is some $\delta>0$ such that $|f(x)|<\epsilon/c$ whenever $|x-x_\star| < \delta$ for $x\in A$. If $|x-x_\star|<\delta$, then we have
$$
|g(x)f(x)| = |g(x)|\cdot|f(x)| \le c\cdot |f(x)| < \epsilon.
$$
As $\epsilon > 0$ was arbitrary, $\lim_{x\to x_\star}g(x)f(x)$ exists and is equal to $0$.

Remark. There are equivalent ways to prove a limit is equal to something over $\mathbb R^n$, one using sequences, and one using $\epsilon$ and $\delta$. The formulation of this using the sequence approach would proceed in much the same way:
Let $\epsilon > 0$ be a challenge number. We will show that for any sequence $(x_n)$ such that $x_n\to x_\star$, then $g(x_n)f(x_n)\to 0$. Since $f(x)\to 0$ as $x\to x_\star$, there is some integer $N$ so that for all integers $n \ge N$, we have $|f(x_n)| \le \epsilon/c$. If $n \ge N \dots$,
I will let you finish the proof.
