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Math Help - Help with Proofs

  1. #1
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    Question Help with Proofs

    I'm taking a new class and instructor isn't very helpful . These are probably very simple, but I'm lost.

    1) If an-->L, then |an|-->|L|

    2) Prove that lim(an-bn) = lim(an)-lim(bn), provided lim(an) and lim(bn) exist

    3) Give an example where lim(an) and lim(bn) do not exist, but lim(an+bn) exists.

    4) If {an} is a bounded sequence and if {bn} is a sequence converging to 0, then {anbn} converges to 0.


    I have tons of questions, but I'll start with these.

    Help is extremely appreciated!
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  2. #2
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    Quote Originally Posted by abradfo1 View Post
    I'm taking a new class and instructor isn't very helpful . These are probably very simple, but I'm lost.

    1) If an-->L, then |an|-->|L|

    2) Prove that lim(an-bn) = lim(an)-lim(bn), provided lim(an) and lim(bn) exist

    3) Give an example where lim(an) and lim(bn) do not exist, but lim(an+bn) exists.

    4) If {an} is a bounded sequence and if {bn} is a sequence converging to 0, then {anbn} converges to 0.
    The secret to #1 is \left| {\left| {a_n } \right| - \left| L \right|} \right| \leqslant \left| {a_n  - L} \right|.

    For #4 Suppose that \left( {\forall n} \right)\left[ {\left| {a_n } \right| \leqslant A} \right] then \left| {a_n b_n } \right| = \left| {a_n } \right|\left| {b_n } \right| \leqslant A\left| {b_n } \right|
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by abradfo1 View Post
    I'm taking a new class and instructor isn't very helpful . These are probably very simple, but I'm lost.

    1) If an-->L, then |an|-->|L|

    2) Prove that lim(an-bn) = lim(an)-lim(bn), provided lim(an) and lim(bn) exist

    3) Give an example where lim(an) and lim(bn) do not exist, but lim(an+bn) exists.

    4) If {an} is a bounded sequence and if {bn} is a sequence converging to 0, then {anbn} converges to 0.


    I have tons of questions, but I'll start with these.

    Help is extremely appreciated!
    2. Since \{a_n\}\to a, \exists~N_1 such that n>N_1 implies that |a_n-a|<\frac{\epsilon}{2}. Similarly, since \{b_n\} \to b, \exists~N_2 such that n>N_2 implies that |b_n-b|<\frac{\epsilon}{2}.

    Let N=\max\{N_1,N_2\}. Thus, when n>N, |(a_n-b_n)-(a-b)| = \underbrace{|(a_n-a)+(b-b_n)| \leq |a_n-a|+|b-b_n|}_{triangle~inequality} < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.

    Therefore \lim(a_n-b_n)=a-b = \lim a_n-\lim b_n.

    3. Consider a_n = 1+(-1)^n and b_n=1-(-1)^n.
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  4. #4
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    Thanks so much. I just had no idea where to start! Thank you!
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