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    only one for now but I'm revising so there's bound to be more. Just starting with proving limits of sequences, although this question is non-specific and probably blindingly obvious. For a step of the proof there is a line:

    \frac{n^2 - 5}{n^3 - 50} < \frac{2}{n}

    but i can;t quite see where it comes from. I know you need the top LHS<top RHS andthe bottoms LHS bottom RHS for the LHS < RHS.

    I tried n^2 - 5 &lt; 2n^2 but then

    n^3 - 50 is not &gt; n^3.

    any help appreciated and I'll try and put edit it into latex
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    Well, it's certainly not true for all n. n = 4, for example.

    Can you prove that n^2 - 5 < 2(n^3 - 50)/n for all sufficiently large values of n?
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    oh yeah, for n>= 5 it works, i forgot that bit
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    (Original post by thestudent)
    I know you need the top LHS<top RHS andthe bottoms LHS bottom RHS for the LHS < RHS.
    In fact, this isn't true at all. If you can prove this then you've proved that LHS < RHS, but sometimes it's not possible; e.g. 8/13 < 5/6. Certainly 13 > 6, but we don't have 8 < 5.
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    (Original post by davidac)
    What the **** is this ****? Mate, just show me the question.
    good point:

    prove the limit of \frac{n^2 - 5}{n^3 - 50} = 0
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    (Original post by thestudent)
    good point:

    prove the limit of \frac{n^2 - 5}{n^3 - 50} = 0
    Divide top and bottom by n^2. Or prove it's <2/n with the method I gave you above.
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    (Original post by generalebriety)
    In fact, this isn't true at all. If you can prove this then you've proved that LHS < RHS, but sometimes it's not possible; e.g. 8/13 < 5/6. Certainly 13 > 6, but we don't have 8 < 5.
    oh yeah... i think I meant, just generally when trying to find a fraction that is greater than another, that's a surefire way, and the way in which i think this one has been found
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    (Original post by thestudent)
    oh yeah... i think I meant, just generally when trying to find a fraction that is greater than another, that's a surefire way, and the way in which i think this one has been found
    That's my point. It's a way, but it's not a surefire way. It does mess up sometimes.
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    ok, so its proving using the whole definition of a sequence thing. So we have for epsilon >0
    where An is the sequence
    |An-l| < epsilon
    |\frac{n^2 - 5}{n^3 - 50} - 0| < epsilon. this can't be solved to get n. so we find a sequence which An is less than. This sequence can be 2/n. I can see An is less than 2/n but how can I get to choosing 2/n (or any other sequence is another sequence), I think that was my question.
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    (Original post by thestudent)
    ok, so its proving using the whole definition of a sequence thing. So we have for epsilon >0
    where An is the sequence
    |An-l| < epsilon
    |\frac{n^2 - 5}{n^3 - 50} - 0| < epsilon. this can't be solved to get n. so we find a sequence which An is less than. This sequence can be 2/n. I can see An is less than 2/n but how can I get to choosing 2/n (or any other sequence is another sequence), I think that was my question.
    I don't understand what you're asking. You can prove that \left|\dfrac{n^2 - 5}{n^3 - 50} - 0\right| can be made arbitrarily small by either simply dividing the top and bottom of the fraction by n^2 (and therefore not changing it) or by comparing it with 2/n as I did above.
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    Can you use L'Hospital's?
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    L'Hopital seems to be labouring the point, especially when simply dividing through by n^2 as general suggested should have killed this thread ages ago :p:
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    Edit function is screwing up for me.

    I suppose you could jusr write "result is trivial by L'Hopital"
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    You could, but you probably won't get any marks. You need to be able to answer questions like this from first principles to have any hope of proving L'Hopital, so using L'Hopital is (indirectly) circular reasoning.
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    I've figured out what I wanted to get in the first place (I think). Simple as 2/n was chosen by n^2 - 5 < n^2 and n^3 - 50 > n^3/2. Thanks for the input thought, even with the obvious display of lack of mathematical intuitiveness on my behalf.
 
 
 
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