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amyna

  • one year ago

find the limit of the sequence n-1/n

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  1. amyna
    • one year ago
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    you take the limit as n approaches infinity

  2. amyna
    • one year ago
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    but i don't know how to start

  3. zepdrix
    • one year ago
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    Is it \(\large\rm \frac{n-1}{n}\) or \(\large\rm n-\frac{1}{n}\) ?

  4. amyna
    • one year ago
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    the first one

  5. zepdrix
    • one year ago
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    Oh limit of the sequence? not the series? Ok that makes things a little easier then :d

  6. zepdrix
    • one year ago
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    \[\large\rm \lim_{n\to\infty}\frac{n-1}{n}\quad=\quad \lim_{n\to\infty}\frac{n}{n}-\frac{1}{n}\quad=\quad \lim_{n\to\infty}1-\lim_{n\to\infty}\frac{1}{n}\]You could break it up like this I suppose. That seems like a good start :o

  7. amyna
    • one year ago
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    ok i will try this approach, can i also multiply both sides by n ?

  8. zepdrix
    • one year ago
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    No no let's not try anything fancy like that :) We don't actually have any "sides" here. It's just this guy \(\large\rm \lim\limits_{n\to\infty}\frac{n-1}{n}\) and we're rewriting him in a way that's easier to handle. Our limit depends on n, So in the same way that you can't factor n `out of the limit`, Example:\(\large\rm \lim\limits_{n\to\infty}2\frac{1}{n}\ne\frac{1}{n}\lim\limits_{n\to\infty}\frac{1}{n}2\) (Can't pull the n stuff out like that) we also don't want to introduce new n stuff into the limit! :o

  9. zepdrix
    • one year ago
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    Ah typo in my example :c woops

  10. zepdrix
    • one year ago
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    I was trying to say that you can't factor the 1/n outside of the limit. Was just trying to give an example why you don't want to multiply any n's into the limit.

  11. amyna
    • one year ago
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    ok got it

  12. zepdrix
    • one year ago
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    \[\large\rm =\quad \lim_{n\to\infty}1-\lim_{n\to\infty}\frac{1}{n}\]But if we were able to break it down into these two limits, these should be manageable. You don't remember that second limit? :o

  13. amyna
    • one year ago
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    then how would i solve it after i break the function up? sorry it has been long since i learned limits so i don't really remember how to actually solve it.

  14. zepdrix
    • one year ago
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    I applied a limit rule to break it into two separate limits. As `n approaches infinity`, what does the number 1 approach?

  15. amyna
    • one year ago
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    infinity as well ?

  16. zepdrix
    • one year ago
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    careful! we're not adding up a bunch of 1's. We're just looking at that number 1 as n grows and grows. 1 is a constant. It's unchanging. So as n gets larger and larger, 1 approaches 1. Hopefully that makes some sense :o

  17. zepdrix
    • one year ago
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    But for the other part... As `n approaches infinity`... let's look at a really big n to see what's going on.\[\large\rm \lim_{n\to\infty}\frac{1}{n}\approx\frac{1}{12bajillion}\]I plugged in a really really big number for n, letting it get "close" to infinity.

  18. zepdrix
    • one year ago
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    This is one of those limits that you really want to remember :) But oh well hehe. So do you notice anything about this number? How about a number you can plug into your calculator, like 1/99999?

  19. amyna
    • one year ago
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    so how would i show the necessary work for this problem since it approaches 1?

  20. zepdrix
    • one year ago
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    You do these simple algebra steps,\[\large\rm \lim_{n\to\infty}\frac{n-1}{n}\quad=\quad \lim_{n\to\infty}\frac{n}{n}-\frac{1}{n}\quad=\quad \lim_{n\to\infty}1-\frac{1}{n}\]And then, at this point in your class, your teacher probably wants you to state what rule you're using here. \[\large\rm =\lim_{n\to\infty}1-\lim_{n\to\infty}\frac{1}{n}\]Justify that step by saying something like... `The limit of a difference is equivalent to the difference of limits`. Or whatever the name of that limit rule is, I can't remember. \[\large\rm =1-0\]\[\large\rm =1\]I think that's all you need to do :o I don't think you need to say anything special about the 1/n limit. That's a pretty common one. I could be wrong though.

  21. amyna
    • one year ago
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    Okay this makes so much more sense now! Thank You for your help!

  22. zepdrix
    • one year ago
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    yay team \c:/

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