**Definition**: A sequence $ \{a_n\} $ has limit $ L $, where $ L $ is a given real number, if for each $ \epsilon > 0 $ there is an integer $ N \ge 1 $ such that

$ |a_n-L| < \epsilon $ for all integers $ n \ge N $.

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Problem: (Stanford Math 51H) Prove that a sequence $ \{a_n\} $ cannot have more than one limit.

Solution: This is logically true, as usual, but a rigorous argument is much more fun.

Suppose $ \lim a_n = L_1 $ and $ \lim a_n = L_2 $. Letting $ \epsilon = \frac{|L_1-L_2|}{2} $, we know there exists $ N_1, N_2 $ such that

$ |a_n-L_1| < \epsilon $ for $ n \ge N_1 $

$ |a_n-L_2| < \epsilon $ for $ n \ge N_2 $.

So the above two inequalities are true for $ n \ge \max(N_1, N_2) $. Adding the two inequalities together, we get

$ |a_n-L_1|+|a_n-L_2| < 2\epsilon = |L_1-L_2| $.

However, by the triangle inequality we know $ |a|+|b| \ge |a-b| $ (by setting $ b = -b $ is the usual one), so

$ |a_n-L_1|+|a_n-L_2| \ge |L_2-L_1| = |L_1-L_2| $.

Then

$ |L_1-L_2| < |L_1-L_2| $,

a contradiction. Hence $ \{a_n\} $ can have at most one limit. QED.

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Comment: The real definition of the limit is almost always complete overlooked in a regular calculus course (i.e. Calc AB and BC). But it's pretty much the foundation of all of calculus so it is really quite nice to know and work with.

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Practice Problem: (Stanford Math 51H, Sandwich Theorem) If $ \{a_n\}, \{b_n\} $ are given convergent sequences with $ \lim a_n = \lim b_n $, and if $ \{c_n\} $ is any sequence such that $ a_n \le c_n \le b_n \forall n \ge 1 $, prove that $ \{c_n\} $ is convergent and $ \lim c_n = \lim a_n = \lim b_n $.

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