# Thread: An interesting limit

1. ## 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

2. 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 $\frac{f'(a)}{g'(a)}$

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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 !

3. ah ok i see what went wrong there. lol.