## Thursday, November 23, 2006

### Limitten. Topic: Real Analysis.

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 $\lim c_n = \lim a_n = \lim b_n$.