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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 $$-M \cdot 0 \le \lim_{x \to 0} \ g(x)h(x) \le M \cdot 0 \text{?} $$

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Will need absolute values in some places. Certainly can say $-M|g(x)| \le g(x)h(x)\le M|g(x)|$. Then if one has already proved the general "squeezing" principle, you can finish more or less like you did. –  André Nicolas Nov 17 '11 at 23:43
    
@AndréNicolas: Or $$ -M \cdot \left|\lim_{x \to 0} \ g(x) \right| \le \left|\lim_{x \to 0} \ g(x)h(x)\right| \le M \cdot \left|\lim_{x \to 0} \ g(x) \right|,$$ right? –  raxacoricofallapatorius Nov 18 '11 at 1:55
    
I would not say so. First comes what I wrote. For your line as second line, we need reference to squeezing theorem, or separate proof. –  André Nicolas Nov 18 '11 at 2:28
    
@AndréNicolas: So (bear with me) where am I going wrong here: $$\left|\lim\limits_{x \to 0} \ g(x)h(x)\right| \le \left|\lim\limits_{x \to 0} \ M \cdot g(x) \right|$$ $$\left|\lim\limits_{x \to 0} \ g(x)h(x)\right| \le \left| M \cdot \lim\limits_{x \to 0} \ g(x) \right|$$ and thus $$ -M \cdot \left|\lim\limits_{x \to 0} \ g(x) \right| \le \left|\lim\limits_{x \to 0} \ g(x)h(x)\right| \le M \cdot \left|\lim\limits_{x \to 0} \ g(x) \right|\text{ ?}$$ –  raxacoricofallapatorius Nov 18 '11 at 4:15
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Sorry for the delay, I was out. You are using an intuitively very reasonable property of limits. But part of the point early on in an analysis course is that you prove what seems intuitively reasonable. Continuous functions, for example, can be a lot weirder than the smooth curves of our imagination. Later, when you know that you could prove something easily, you can have the luxury of not doing every grungy detail. –  André Nicolas Nov 18 '11 at 5:56
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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 $\varepsilon$.

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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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Are you sure? I think I'm just factoring out a constant. –  raxacoricofallapatorius Nov 17 '11 at 23:47
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@raxacoricofallapatorius Your argument proves that if the limit exists, then it is $0$. Your argument does not prove that the limit exists. –  Michael Hardy Nov 18 '11 at 0:09
    
Is 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 $c\cdot\lim_{x \to 0} f(x)$)? –  raxacoricofallapatorius Nov 18 '11 at 0:28
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