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\small \normalsize %% dumb, but have to do this for the prev to work
Jeremy Osterhouse \\
\today
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{\bf Problem.} Prove the Pinching Theorem: \emph{Suppose that for all sufficiently large n,}
\[
a_n \leq b_n \leq c_n.
\]
\emph{If $\lim{a_n} = \lim{c_n} = L$, then $\lim{b_n}=L$.}

{\bf Solution.} To prove this, we will use the definition of the limit of a sequence. Because we know that $\lim{a_n} = \lim{c_n} = L$, we know by the definition of the limit of a sequence that,
\[
L-\epsilon_a < a_n < L+\epsilon_a
\]
for any $\epsilon_a$ with all sufficiently large $n$ and that,
\[
L-\epsilon_c < c_n < L+\epsilon_c
\]
for any $\epsilon_c$ with all sufficiently large $n$.

Because we know that $a_n<c_n$, let $\epsilon = max\{\epsilon_a,\epsilon_c\}$. Now we get
\[
L-\epsilon < a_n \leq c_n < L + \epsilon
\]
But we know that $a_n \leq b_n \leq c_n$. For any $\epsilon$, then there exists an $\epsilon_b$ such that for sufficiently large $n$,
\[
L-\epsilon \leq L-\epsilon_b < a_n \leq b_n \leq c_n < L+\epsilon_b \leq L+\epsilon.
\]
Removing the unneccesary terms of this inequality we get,
\[
L-\epsilon < b_n < L+\epsilon
\]
for any $\epsilon$ with all sufficiently large $n$. This is equivalent to saying that $\lim{b_n}=L$.
\hfill$\square$

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