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Can anyone prove Squeeze theorem for me, which says:

(1):if $f(x)< g(x) < h(x)$ for all $x$ near $a$, and $\lim\limits_{x\to a}f(x) = \lim\limits_{x \to a}h(x) = L$

then $\lim\limits_{x \to a}g(x)=L$.

(2): If $f$ dominates $g$, that is, ($f(x) \geq g(x)$) near $x=a$ and $g(x) \to \infty$ as $x \to \infty$ then so does $f(x)$

Thank you.

ABC
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LoveMath
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3 Answers3

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Hints:

For 1: choose $\delta_f$ and $\delta_h$ such that $|x - a| < \delta_f$ implies $|f(x) - L| < \epsilon$ and $|x - a| < \delta_h$ implies $|h(x) - L| < \epsilon$. Let $\delta$ be the smaller of $\delta_f$ and $\delta_h$. Now if $|x - a| < \delta$ then both $|f(x) - L|$ and $|h(x) - L|$ are less than $\epsilon$. Use that, along with $f(x) < g(x) < h(x)$, and you will be able to show that $|g(x) - L| < \epsilon$. This will prove that $g(x) \to L$ as $x \to a$.

For 2: If $f$ dominates $g$ near $a$ then I suspect you want to take the limit as $x \to a$ right? I'll leave this one to you, if you just write down the definition of the limit for $g$ you should be able to turn that into the definition for a limit of $f$.

Jim
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We have $$|g(x)-L|\leq \max(|f(x)-L|,|h(x)-L|)$$ Let $\epsilon>0$ then there's $\delta'>0$, if $|x-a|\leq \delta'$ we have $|f(x)-L|<\epsilon$ and there's $\delta''>0$, if $|x-a|\leq \delta''$ we have $|h(x)-L|<\epsilon$

Take $\delta=\min(\delta',\delta'')$ and if $|x-a|\leq \delta$ we have $|g(x)-L|<\epsilon$

For the second result, let $A>0$ then there's $\delta>0$, if $|x-a|\leq \delta$ we have $f(x)\geq g(x)>A$

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If $$f<g<h \ \forall \ x\in \Bbb R$$ and $$\lim_{x\to a}f=\lim_{x\to a}h=L$$

We have $g$ between $f \ and \ h$ so the $g$ must be $L\Big|_{x\to a}$ , otherwise $f<g<h$ would not hold true.

In other words $f<g<h$ squeeze the $g$ between both $f,h$ when $f$ and $g$ tend to $L$

ABC
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    A nice intuitive explanation, but not a proof. – Jim Apr 11 '13 at 07:12
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    I think he intended a proof by contradiction, though not written rigorously. Assume the contradiction, let $g(x)$ go to $M$ instead of $L$. Then if $M>L$, then the upper bound condition will be violated, if $M<L$, then the lower bound condition will be violated. – dineshdileep Apr 11 '13 at 08:07
  • "We have $h$ between $f$ and $h$" ? Is it? or, "We have $g$ between $f$ and $h$." – learner Apr 12 '13 at 13:36