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    If anyone could just show me how to do this limit it would be much appreciated. My mind's just gone completely blank

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    Eek! Wouldn't know where to start!! I would, if I had to have a stab, use the MacLaurin/Taylor series, but that's probably the entirely wrong thing to do...

    Just forget about me... I've only covered AS :p:
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    (Original post by henryt)
    Eek! Wouldn't know where to start!! I would, if I had to have a stab, use the MacLaurin/Taylor series, but that's probably the entirely wrong thing to do...

    Just forget about me... I've only covered AS :p:
    I've tried to get Maple to do the series for x.tan(1/x) but it just says Error, so I don't think that's how to go about it annoyingly
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    The limit is the same as tan(y)/y as y goes to zero. But this is

    (sin(y)/y)/cos(y)

    And we know that sin(y)/y tends to 1 as y tends to zero (use the maclaurin series, although it's not entirely rigorous unless you know all the theory). Also, cos(y) tends to 1 as y tends to zero so the required limit is 1.
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    (Original post by mikesgt2)
    The limit is the same as tan(y)/y as y goes to zero. But this is

    (sin(y)/y)/cos(y)

    And we know that sin(y)/y tends to 1 as y tends to zero (use the maclaurin series, although it's not entirely rigorous unless you know all the theory). Also, cos(y) tends to 1 as y tends to zero so the required limit is 1.
    I agree with that.
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    let u = 1/x
    The limit can be written as
    1/u (sinu/cosu) while u tends to 0
    = (sinu)/u because cos u tends to 1
    = 1 well I think that's a standard result you need to learn
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    (Original post by olliemccowan)
    If anyone could just show me how to do this limit it would be much appreciated. My mind's just gone completely blank


    let w=1/x

    we want lim w->0 tanw/w = lim->0 sec^2w =1 Using L'hopitals Rule.
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    (Original post by mikesgt2)
    The limit is the same as tan(y)/y as y goes to zero. But this is

    (sin(y)/y)/cos(y)

    And we know that sin(y)/y tends to 1 as y tends to zero (use the maclaurin series, although it's not entirely rigorous unless you know all the theory). Also, cos(y) tends to 1 as y tends to zero so the required limit is 1.
    Thanks I've just got to get into the swing of things so I can pick up on the little things like that which make the question so simple.

    Thanks everyone for your help
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    the taylor eaxpansion of tan 1/x is contains infinite terms starting
    1/x + 1/3x^3 + ...
    Mulitiply this by x and you get
    1 + 1/3x^2 + ...
    all terms other than 1 go to 0 as x goes to infinity so by Algebra of limits the sum goes to 1
 
 
 
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