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Math Help - Limit fraction.

  1. #1
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    Limit fraction.

    Hi,

    I don't see how this leap is made.
    \lim_{x \to y} \frac{x^n-y^n}{x-y}

    equals

    \lim_{x \to y} x^{n-1} + x^{n-2}y + . . . + xy^{n-2} + y^{n-1}

    and this is even more confusing.

    =ny^{n-1}
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by craigmain View Post
    Hi,

    I don't see how this leap is made.
    \lim_{x \to y} \frac{x^n-y^n}{x-y}

    equals

    \lim_{x \to y} x^{n-1} + x^{n-2}y + . . . + xy^{n-2} + y^{n-1}

    and this is even more confusing.

    =ny^{n-1}
    Your problem is not Calculus, but Algebra!

    Its an algebraic identity x^n-y^n = (x-y)(x^{n-1} + x^{n-2}y + . . . + xy^{n-2} + y^{n-1}). You should know this basic identity....
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  3. #3
    Moo
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    Hello,

    See here for a way to prove it : http://www.mathhelpforum.com/math-he...484-k-b-k.html

    As for the initial limit, you can recognize a form of the difference quotient :
    \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=f'(x)

    Just substitute x=y+h, with f(t)=x^n, in what you've been given to get the above.
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  4. #4
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    Quote Originally Posted by craigmain View Post
    Hi,

    I don't see how this leap is made.
    \lim_{x \to y} \frac{x^n-y^n}{x-y}

    equals

    \lim_{x \to y} x^{n-1} + x^{n-2}y + . . . + xy^{n-2} + y^{n-1}

    and this is even more confusing.

    =ny^{n-1}
    It might be easier is you see some numbers for n

     <br />
x^2-y^2 = (x-y)(x+y)<br />
    x^3-y^3 = (x-y)(x^2+xy+y^2)
     <br />
x^4-y^4 = (x-y)(x^3 + x^2y+xy^2+y^3)<br />
     <br />
x^5-y^5 = (x-y)(x^4 + x^3y+x^2y^2+xy^3+y^4)<br />

    Notice the number of terms is the second () and the power on the LHS (2 and 2, 3 and 3, 4 and 4, 5 and 5 etc), and that when x \to y all the terms are the same.
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  5. #5
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    Quote Originally Posted by Moo View Post
    Hello,

    See here for a way to prove it : http://www.mathhelpforum.com/math-he...484-k-b-k.html

    As for the initial limit, you can recognize a form of the difference quotient :
    \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=f'(x)

    Just substitute x=y+h, with f(t)=x^n, in what you've been given to get the above.
    Using a geometric series to prove that - I like that!
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  6. #6
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    And, finally, taking x= y in x^{n-1}+ x^{n-2}y+ \cdot\cdot\cdot+ xy^{n-2}+ y^{n-1} makes every term equal to y^{n-1} and, since there are n terms, the sum is ny^{n-1}.
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