Kinda Urgent. Proof Wriiting. Squences

  • Thread starter psycho2499
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Homework Statement


Prove that if X[n]->L, X[n] doesn't equal 0 for all n, L doesn't equal 0 and Y[n]->M, then (Y[n]/X[n])->(M/L).


Homework Equations


They only give you the squeeze theroem and that if X[n] converges then it's limit is unique. O and the definition. A sequence of reals has limit L iff for every epsilon>0 there exists a natural number N such that if n>N, then |X[n]-L|<epsilon



The Attempt at a Solution


The problem lies in I don't even know where to begin. I know to suppose the antecendent and i think that i let epsilon>0 and choose N belonging to the set of Naturals such that N is greater then some number but I have no idea what it would be. Usually in the examples and previous problems it would be something like, for some real number C, C*epsilon but I don't see a strategy like that taking me anywhere.
 

Answers and Replies

  • #2
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The idea is to start with the conclusion (Y[n]/X[n])->(M/L). If this is indeed true, then apply the definition to this limit. See if you can somehow bind the relevant expression resulting from this last limit by epsilon using the assumptions.
 
  • #3
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The idea is to start with the conclusion (Y[n]/X[n])->(M/L). If this is indeed true, then apply the definition to this limit. See if you can somehow bind the relevant expression resulting from this last limit by epsilon using the assumptions.

Could you clarify this at all. What i sounds like you are saying to me is that I should be supposing the consequent but that is uber taboo. I mean I could suppose the my antecedent and then bind |Y[n]/X[n]-M/L|<epsilon but I'd have to get there algebraically first.
 
  • #4
1,101
3
Sorry, I should have said to work from the assumptions to show |Y[n]/X[n]-M/L|<epsilon, which is what we need. Clearly, |Y[n]/X[n]-M/L| = |Y[n]*L - X[n]M| / |X[n]*L|. Now we know that |Y[n] - M| < a for any a > 0 and |X[n] - L| < b for any b > 0. Can you factor the numerator of |Y[n]*L - X[n]M| / |X[n]*L| to get an equivalent expression involving both |Y[n] - M| and |X[n] - L|? From there, we need to pick suitable a and b to bind the resulting expression further.
 
  • #5
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Ok this isn't so urgent any more but my teacher loves to just stare blankly at me like I'm stupid when I ask questions so I'd like to continue here with anyone who will help. I got to |(Y[n]*L-M*L+M*L-X[n]*M)/X[n]*L|<epsilon in which i then factored |(L*(Y[n]-M)-M*(X[n]-L))/X[n]*L|<epsilon. I got the feeling from what you were saying earlier that I should try sandwhich theroem and this is where my math and logic get shotty I'm pretty certain but my teacher just is to vague. I stated that there exists a,b belonging to the set of Reals such that |X[n]-L|<b and |Y[n]-M|<a. But Now that I'm reviewing my work it looks like I have no idea how to even begin to find the bounds for applying squeeze or then applying sqeeze.
 

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