I've seen several variations on the exercise to show that $\lim\limits_{x \to 0} \ g(x) h(x) = 0$ when $\lim\limits_{x \to 0} \ g(x)=0$ and $|h(x)| \le M$, all of which rely on the $\varepsilon$-$\delta$ definition of the limit of a function. However it's not clear to me why it is necessary to use the definition to show this. What's wrong with writing $ -M \cdot \lim_{x \to 0} \ g(x) \le \lim_{x \to 0} \ g(x)h(x) \le M \cdot \lim_{x \to 0} \ g(x) ,$ and thus $-M \cdot 0 \le \lim_{x \to 0} \ g(x)h(x) \le M \cdot 0 \text{?} $
Showing that $\lim \limits_{x \to 0}g(x)h(x)=0$ when $\lim \limits_{x \to 0}g(x)=0$ and $\left|{h(x)}\right|\le M$
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0@A$n$dréNicolas: Yes, this is the tough part of wrapping one's head around analysis (or any "real math"), coming from a more "applied" cast of mind. I have [another question](http://math.stackexchange.com/questions/83658/proving-differentiability-by-demonstrating-what-the-derivative-is) where this is explicitly the issue. – 2011-11-19
2 Answers
A conventional squeezing argument might start with this: $ -M g(x) \le g(x)h(x) \le M g(x),$ with no "$\lim$" in it, and then cite the fact that $\lim\limits_{x\to0} g(x)=0$ to conclude that $\lim\limits_{x\to0} g(x)h(x) = 0$. In this case, you know that $-M\le h(x)\le M$. You cannot go from there to the conclusion that $-Mg(x) \le g(x)h(x) \le Mg(x)$ unless you know that $g(x)\ge 0$, and that was not among your assumptions. If you did have that hypothesis, then squeezing like this would work.
Since $\lim\limits_{x\to0} g(x)=0$, you can say that given $\varepsilon>0$, there must exists $\delta>0$ so small that if the distance between $x$ and $0$ is less than $\delta$ but not $0$, then the distance between $g(x)$ and $0$ is less than $\varepsilon/M$. Then the distance between $h(x)g(x)$ and $0$ must be less than $\varepsilon$.
This inequality assumes that the limit exists, which would need to be proved some way before you can assert the inequality.
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0Is it not always true that when $\lim_{x \to 0} f(x)$ exists, $\lim_{x \to 0} \ c\cdot f(x)$ also exists (and is $c\cdot\lim_{x \to 0} f(x)$)? – 2011-11-18