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- Thread starter goody
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- Mar 1, 2012

- 658

actually, factoring out $(x-a)$ in the numerator yields ...Good evening! Could anybody help me with this limit?

View attachment 10126

I have a problem when there are two variables. The only thing I did was that:

View attachment 10129,

but I don't know if it was helpful. Thank you!

$\dfrac{(x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ... + xa^{n-2} + a^{n-1} - na^{n-1})}{(x-a)^2}$

I played with the algebra to see if I could get the other $(x-a)$ factor in the denominator to divide out ... no joy, yet.

So, starting again ...

$\displaystyle \lim_{x \to a} \dfrac{(x^n-a^n) - na^{n-1}(x-a)}{(x-a)^2}$

using L'Hopital ...

$\displaystyle \lim_{x \to a} \dfrac{nx^{n-1} - na^{n-1}}{2(x-a)}$

$\displaystyle \lim_{x \to a} \dfrac{n(x^{n-1} - a^{n-1})}{2(x-a)}$

L'Hopital again ...

$\displaystyle \lim_{x \to a} \dfrac{n(n-1)(x^{n-2})}{2} = \dfrac{n(n-1)(a^{n-2})}{2}$

- Jan 30, 2018

- 428

No, there aren't. There areGood evening! Could anybody help me with this limit?

View attachment 10126

I have a problem when there are two variables.

No. [tex]x^n- a^n[/tex] is NOT [tex](x- a)(x^{n-1}- a^{n-1}[/tex]. If you multiply the right side of that you get [tex]x^n- a^{n-1}x^n- ax^{n-1}+ a^n[/tex]. [tex]x^n- a^n= (x- a)(x^{n-1}+ ax^{n-2}+ \cdot\cdot\cdot+ a^{n-2}x+ a^{n-1}[/tex].The only thing I did was that:

View attachment 10129,

but I don't know if it was helpful. Thank you!

- Mar 1, 2012

- 658

Quoting the last part of your post ...

No. $x^n- a^n$ is NOT $(x- a)(x^{n-1}- a^{n-1})$.

If you multiply the right side of that you get $x^n- a^{n-1}x^n- ax^{n-1}+ a^n$.

$x^n- a^n= (x- a)(x^{n-1}+ ax^{n-2}+ \cdot\cdot\cdot+ a^{n-2}x+ a^{n-1})$.

- Jan 30, 2018

- 428

Thank you. I am on entirely too many boards with too many different protocols!