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Any help in solving the following problem would be greatly appreciated:

Let $f, g_1, g_2$ be functions from $\mathbb R$ to $\mathbb R$, with $g_1(x) \leq f(x) \leq g_2(x)$, for all $x \in \mathbb R$. Suppose that, for some $p \in \mathbb R$, we have $\lim_{x \rightarrow p} g_1(x) = \lim_{x \rightarrow p} g_2(x) = c$. Show that $ \lim_{x\rightarrow p} f(x) = c,$ as well.

I've spent hours on it and haven't come up with anything worth posting.

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What do you mean by $\lim_{x \rightarrow p} > g_1(x) $? – Hawk Nov 30 '12 at 19:57
$\lim_{x \rightarrow p} > c$ ? – HipsterMathematician Nov 30 '12 at 20:02
@sizz: I suspect that the $>$ is just a typo. – Brian M. Scott Nov 30 '12 at 20:03
isn't it the "squeeze theorem" for limits?if $g_1$ and $g_2$ have limit equal $c$ then $f$ has limit equal $c$? – HipsterMathematician Nov 30 '12 at 20:05
@Charlie, in that case – Hawk Nov 30 '12 at 20:08
up vote 4 down vote accepted

HINT: Let $\epsilon>0$. There are real numbers $\delta_1,\delta_2>0$ such that $$|g_1(x)-c|<\frac{\epsilon}2\quad\text{whenever}\quad 0<|x-p|<\delta_1$$ and $$|g_2(x)-c|<\frac{\epsilon}2\quad\text{whenever}\quad 0<|x-p|<\delta_2\;.$$

Now show that

$$|f(x)-c|<\epsilon\quad\text{whenever}\quad 0<|x-p|<\min\{\delta_1,\delta_2\}\;.$$

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@BrainMScott Thank you! The procedure seems simple when one examines it. – Neil Dec 3 '12 at 18:29
@Neil: You’re welcome. Experience helps in seeing where to start, but I agree that the basic idea isn’t hard. – Brian M. Scott Dec 3 '12 at 21:07

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