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    Hi,

    Are my answers correct?

    Q1) lim x --> 9
    x - 9 / \sqrt x -3
    My answer = 9

    Q2) lim a -->-5
    a^3 + 3a^2 -10a /( a+5)
    My answers: a= 2 and a=0

    Q3) lim a--> 0
    (a^1/3) / (a^2 - 2a^1/3 )
    Thats (a power third) divided by (a squared - 2a to third)
    My answer = -0.5

    thanks!
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    (Original post by dream123)
    Hi,

    Are my answers correct?

    Q1) lim x --> 9
    x - 9 / \sqrt x -3
    My answer = 9

    Q2) lim a -->-5
    a^3 + 3a^2 -10a /( a+5)
    My answers: a= 2 and a=0

    Q3) lim a--> 0
    (a^1/3) / (a^2 - 2a^1/3 )
    Thats (a power third) divided by (a squared - 2a to third)
    My answer = -0.5

    thanks!
    http://www.wolframalpha.com/

    This website solves limits directly. Enter like this "limit as x tends to 9 of (x-9)/(sqrt(x)-3)".

    Btw, q1 is actually 6.
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    (Original post by Don John)
    http://www.wolframalpha.com/

    This website solves limits directly. Enter like this "limit as x tends to 9 of (x-9)/(sqrt(x)-3)".

    Btw, q1 is actually 6.
    I've just used the L'hospital's method of simply differentiation and my answers are correct
    Only problem is, my question has 10 marks-do you think l'hospitals method is work 10 marks? :/
    Thanks
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    (Original post by dream123)
    I've just used the L'hospital's method of simply differentiation and my answers are correct
    Only problem is, my question has 10 marks-do you think l'hospitals method is work 10 marks? :/
    Thanks
    Use the method that you've been taught in lectures. If you don't know, simply email your lecturer and ask what method is usually accepted in this module.
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    (Original post by dream123)
    Are my answers correct?
    I don't think so:


    Q1) lim x --> 9
    x - 9 / \sqrt x -3
    My answer = 9
    How did you get 9 as your answer? I don't think it's right.

    Q2) lim a -->-5
    a^3 + 3a^2 -10a /( a+5)
    My answers: a= 2 and a=0
    What do you mean 'a=...'? You're not trying to find a, you're trying to find the limit of the fraction as a \to 5. There is only one limit, not two, so you definitely shouldn't have 2 limits here.

    Q3) lim a--> 0
    (a^1/3) / (a^2 - 2a^1/3 )
    Thats (a power third) divided by (a squared - 2a to third)
    My answer = -0.5
    I agree with this answer.

    By the looks of it, Q1,2 are L'Hôpital's rule questions, and Q3 just requires you to rearrange the fraction to find the limit.
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    (Original post by nuodai)
    I don't think so:



    How did you get 9 as your answer? I don't think it's right.

    What do you mean 'a=...'? You're not trying to find a, you're trying to find the limit of the fraction as a \to 5. There is only one limit, not two, so you definitely shouldn't have 2 limits here.


    I agree with this answer.

    By the looks of it, Q1,2 are L'Hôpital's rule questions, and Q3 just requires you to rearrange the fraction to find the limit.
    I figured it out, so all is well
 
 
 
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