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Math Help - An interesting limit

  1. #1
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    An interesting limit

    Consider the following limit:

    lim (x^m-a^m)/(x^n-a^n), a is not = 0, m,n are positive integers
    x-->a

    this obviously gives us a 0/0 limit if we use direct substitution, hence I was told to use l'Hopital's rule to evaluate this limit.

    However I ran into a little problem in that if i keep differentiating the top and the bottom, i'll just keep getting a limit in the indeterminate 0/0 form.

    This carries on until the very end when i get

    (m(m-1)(m-2)....(3)(2)(1)x^(m-m)-m(m-1)(m-2)...(3)(2)(1)a^(m-m))
    __________________________________________________ _______
    (n(n-1)(n-2)...(3)(2)(1)x^(n-n)-n(n-1)(n-2)...(3)(2)(1)a^(n-n))

    which is as far as i can see still an indeterminate form.

    So am i right to suggest that this limit cannot be evaluated?

    Any help would be welcome.

    Thanks
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  2. #2
    Moo
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    Hello !

    Here is my point : I hate l'Hospital's rule !

    Do you know the formula of the derivative number : \lim_{x \to a} \frac{f(x)-f(a)}{x-a}=\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}=f'(a) ?

    Consider f(x)=x^m and g(x)=x^n

    The limit can be rewritten this way :

    =\lim_{x \to a} \frac{x^m-a^m}{x^n-a^n} \cdot \frac{x-a}{x-a} (it is allowed to multiply by 1 !)

    =\lim_{x \to a} \frac{x^m-a^m}{x-a} \times \frac{x-a}{x^n-a^n}

    Assuming the limits exist, one can see that this is \frac{f'(a)}{g'(a)}

    --------------------------------------------------------------
    However I ran into a little problem in that if i keep differentiating the top and the bottom, i'll just keep getting a limit in the indeterminate 0/0 form.
    Hey dear, a^m and a^n are constant and their derivative is 0 !
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  3. #3
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    ah ok i see what went wrong there. lol.
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