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This is a simple proof of Real Analysis. Show that $\lim_{x \rightarrow a} f(x)=0$, given $|f(x)|\leq g(x)$, $\forall x\in\mathbb R$ and $\lim_{x \rightarrow a} g(x)=0$.

How do you show that $\lim_{x \rightarrow a} f(x)=0$?

I tried by myself $\forall\epsilon>0\exists\delta>0\forall x\in\mathbb{R}:|x-a|<\delta\implies |g(x)|<\epsilon.$ Since $g(x) \leq f(x)\leq g(x)$ By squeeze theorem. I think $\lim_{x \rightarrow a} g(x)=0$

Charlie Van Basten Øydne
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    You know the squeeze theorem? –  Apr 17 '20 at 20:27
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    Like this $$|lim _{x \rightarrow a} f(x)| ≤ |lim _{x \rightarrow a} g(x)|$$ – Charlie Van Basten Øydne Apr 17 '20 at 20:29
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    $$|lim _{x \rightarrow a} f(x)| ≤ 0$$ $$0≤ lim _{x \rightarrow a} f(x) ≤ 0$$ – Charlie Van Basten Øydne Apr 17 '20 at 20:30
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    What have you tried so far? I would use the squeeze theorem first to show that the limit of $|f(x)|$ must be nonnegative, and since $g$ approaches 0 as $x$ approaches $a$ and $g(x)\geq |f(x)|$, then $|f(x)|$ tends to $0$ as $x$ approaches $a$, and then to show that the limit of the absolute value of a function approaches zero as $x$ approaches $a$ implies the original function approaches zero as $x$ approaches $a$ can be found here: https://math.stackexchange.com/questions/799115/limit-of-the-absolute-value-of-a-function – teddy Apr 17 '20 at 20:31

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Let $\varepsilon>0$, there exists $\delta$ such that $\forall x\in(a-\delta,a+\delta),g(x)<\varepsilon$ because $\lim\limits_{x\rightarrow a}g(x)=0$. Thus $\forall x\in(a-\delta,a+\delta),|f(x)|\leqslant g(x)<\varepsilon$ so that $\lim\limits_{x\rightarrow a}f(x)=0.$

Tuvasbien
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You know $\forall\epsilon>0\exists\delta>0\forall x\in\mathbb{R}:|a-x|<\delta\implies |g(x)|<\epsilon.$ Since $|f(x)|\leq g(x)\leq |g(x)|$ it follows $$\forall\epsilon>0\exists\delta>0\forall x\in\mathbb{R}:|a-x|<\delta\implies |f(x)|\leq |g(x)|<\epsilon.$$

mag
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