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Math Help - complex limit please solve

  1. #1
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    complex limit please solve

    lim ( x^3 (x-3)^5 -64)/(x-4)
    x→4
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  2. #2
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     <br />
lim\frac{x^3(x-3)^5-64}{x-4}<br />

    use l'H˘pital's rule which says lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}

     <br />
lim\frac{3x^2(5)(x-3)^4}{1}<br />

    fill in x as 4

     <br />
lim\frac{3(4)^2(5)(4-3)^4}{1}<br />

     <br />
3(16)(5)(1^4)<br />

     <br />
lim=240<br />

    i think thats right..hope i helped
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  3. #3
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    Quote Originally Posted by diaahb View Post
    lim ( x^3 (x-3)^5 -64)/(x-4)
    x→4
    i think this is a easy but lengthy question. expand the numerator in the form (x^2-a^2)=(x-a)(x+a) as 64=2^3.after some calculations u will find that x-4 appears in the numerator as well. so x-4 gets cancelled. now put the linits and u get the required answer. if the calculations get too lengthy use L'Hospital's rule.
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  4. #4
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    Quote Originally Posted by Pulock2009 View Post
    i think this is a easy but lengthy question. expand the numerator in the form (x^2-a^2)=(x-a)(x+a) as 64=2^3.after some calculations u will find that x-4 appears in the numerator as well. so x-4 gets cancelled. now put the linits and u get the required answer. if the calculations get too lengthy use L'Hospital's rule.

    please write the steps of answer without L'Hospital's rule.
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  5. #5
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    Quote Originally Posted by diaahb View Post
    please write the steps of answer without L'Hospital's rule.
    1. Do not yell at the members of this forum.

    2. You have been given a perfectly clear and legitimate method as to how to solve it using L'Hospital's Rule. Try it yourself and if you are still having trouble, then post your working and we will try to rectify any mistakes.
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  6. #6
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    Quote Originally Posted by diaahb View Post
    please write the steps of answer without L'Hospital's rule.
    Why? L'Hopital's rule is by far the simplest way to do this. If you really refuse to use L'Hopital's rule, then you can multiply out (x-3)^5 to get the numerator is "standard form" then factor out (x-4) (since the numerator is 0 when x= 4, we know it must have such a factor)as Pulock2009 suggested.
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