TITLE: A question about using Squeeze Theorem to solve theoretical convergence question
QUESTION [0 upvotes]: Could you give me some hint how to deal with this question:
Suppose $a_n\le b_n \le c_n$ for almost all n, $b_n\to L$, $c_n-a_n\to 0$.
Prove: $a_n \to L,b_n \to L$.
Well, if $a_n\to a, b_n \to b$ and $c_n-a_n\to 0$, than $a=b$ and from Squeeze Theorem we can conclude a=b=L.
So, we need to prove that this is only possible case.
$a_n$ is bounded above by L, so it could not converge to $+\infty$,
$b_n$ is bounded below by L, so it could not converge to $-\infty$.
if $c_n-a_n\to 0$ it is not possible that $a_n\to $ some finite number and $b_n \to +\infty$ 
or  $b_n\to $ some finite number and $a_n \to -\infty$.
I failed to eliminate the possibility of $a_n \to -\infty$ and $b_n \to +\infty$,
and the possibility of $a_n,b_n$ do not have any limit at all.
Thanks.

REPLY [0 votes]: We have $0\leq b_n-a_n\leq c_n-a_n$.  Hence by the squeeze thm, $b_n-a_n\rightarrow 0$.  Now apply the triangle inequality:  $|a_n-L|\leq |a_n-b_n|+|b_n-L|$.  Everything on the RHS goes to 0, implying a_n converges to L.  The other case is similar.
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