Prove directly from the definition of the limit that if a_n --> L, the, |a_n| --> |L|. (where a_n means a sub n) Solution The definition of the limit says that if the sequence A_n converges to the limit L, then: For every e > 0, there is some N such that for all n > N, | A_n - L | < e. (*) By hypothesis, we know that (*) is true. So, choose some e > 0, some N which satisfies (*), and some n > N. Now, for all x and y, ||x| - |y|| = |x - y|, if x and y are both nonnegative = |(-x) - (-y) = |x - y|, if x and y are both negative = |(-x) - y| < |x - y|, if x is negative and y is nonnegative = |x - (-y)| < |x - y|, if x is nonnegative and y is negative. So, we always have ||x| - |y|| = |x - y|. This means that | |A_n| - |L| | = |A_n - L| < e. Therefore, For every e > 0, there is some N such that for all n > N, | |A_n| - |L| | < e, (**) since whatever e we choose, we can use the same N as we did for (*). But (**) is, by definition, equivalent to the sequence |A_n| converging to |L|. This completes the proof.

1. Prove directly from the definition of the limit that if a_n --> L, the, |a_n| --> |L|. (where a_n
means a sub n)
Solution
The definition of the limit says that if the sequence A_n converges to the limit L,
then: For every e > 0, there is some N such that for all n > N, | A_n - L | < e. (*) By hypothesis,
we know that (*) is true. So, choose some e > 0, some N which satisfies (*), and some n > N.
Now, for all x and y, ||x| - |y|| = |x - y|, if x and y are both nonnegative = |(-x) - (-y) = |x - y|, if x
and y are both negative = |(-x) - y| < |x - y|, if x is negative and y is nonnegative = |x - (-y)| < |x -
y|, if x is nonnegative and y is negative. So, we always have ||x| - |y|| = |x - y|. This means that |
|A_n| - |L| | = |A_n - L| < e. Therefore, For every e > 0, there is some N such that for all n > N, |
|A_n| - |L| | < e, (**) since whatever e we choose, we can use the same N as we did for (*). But
(**) is, by definition, equivalent to the sequence |A_n| converging to |L|. This completes the
proof