# Thread: Irrational Case of a Limit

1. ## Irrational Case of a Limit

If $\displaystyle n\in\mathbb{R}$, find $\displaystyle \displaystyle \lim_{x\to{a}}\frac{x^n-a^n}{x-a}$

Found for $\displaystyle n\in\mathbb{Q}$:
Spoiler:
1. If $\displaystyle n\in\mathbb{N}$, then $\displaystyle \displaystyle \lim_{x\to{a}}\frac{x^n-a^n}{x-a} = \lim_{x\to{a}}\left(\frac{1}{x-a}\right)\left(x-a\right)\sum_{k=0}^{n-1}x^{n-k-1}a^k = \lim_{x\to{a}}\sum_{k=0}^{n-1}x^{n-k-1}a^k = na^{n-1}$
2. If $\displaystyle n\in\mathbb{Q}_{> 0}$, then $\displaystyle n = \frac{p}{q}$, for some $\displaystyle p, q, \in\mathbb{N}_{> 0}$. If $\displaystyle y = x^{\frac{1}{q}}[/Math] and [Math]b = a^{\frac{1}{q}}$, then $\displaystyle \displaystyle \lim_{x\to{a}}\dfrac{x^n-a^n}{x-a} = \lim_{x\to{a}}\frac{x^{\frac{p}{q}}-a^{\frac{p}{q}}}{x-a} = \lim_{x\to{a}}\frac{(x^{\frac{1}{q}})^p-(a^{\frac{1}{q}})^p}{x-a} = \lim_{y\to{b}}\frac{y^p-b^p}{y^q-b^q}$ $\displaystyle = \displaystyle \lim_{y\to{b}}\frac{(y^p-b^p)/(y-b)}{(y^q-b^q)/(y-b)}$ and by (1) this is $\displaystyle \dfrac{pb^{p-1}}{qb^{q-1}} = \frac{p}{q}b^{p-q} = \left(\frac{p}{q}\right)(b^p)^{(1-\frac{q}{p})} = na^{n-1}$.
3. If $\displaystyle n\in\mathbb{Q}_{< 0}$, then let $\displaystyle n = -m$. We have $\displaystyle \displaystyle \lim_{x\to{a}}\dfrac{x^n-a^n}{x-a} = \lim_{x\to{a}}\dfrac{x^{-m}-a^{-m}}{x-a} = \lim_{x\to{a}}\dfrac{(a^m-x^m)/(x^ma^m)}{x-a} = \lim_{x\to{a}}\left(\dfrac{-1}{x^ma^m}\right)\left(\dfrac{x^{m}-a^{m}}{x-a}\right)$ and by (2) this is $\displaystyle \frac{-1}{a^{2m}}\cdot ma^{m-1} = -ma^{-m-1} = na^{n-1}$.

How do you deal with this when $\displaystyle n\not\in\mathbb{Q}$? Or maybe there is a way of doing it on one-go?

2. Originally Posted by TheCoffeeMachine
If $\displaystyle n\in\mathbb{R}$, find $\displaystyle \displaystyle \lim_{x\to{a}}\frac{x^n-a^n}{x-a}$

Found for $\displaystyle n\in\mathbb{Q}$:
Spoiler:
1. If $\displaystyle n\in\mathbb{N}$, then $\displaystyle \displaystyle \lim_{x\to{a}}\frac{x^n-a^n}{x-a} = \lim_{x\to{a}}\left(\frac{1}{x-a}\right)\left(x-a\right)\sum_{k=0}^{n-1}x^{n-k-1}a^k = \lim_{x\to{a}}\sum_{k=0}^{n-1}x^{n-k-1}a^k = na^{n-1}$
2. If $\displaystyle n\in\mathbb{Q}_{> 0}$, then $\displaystyle n = \frac{p}{q}$, for some $\displaystyle p, q, \in\mathbb{N}_{> 0}$. If $\displaystyle y = x^{\frac{1}{q}}[/Math] and [Math]b = a^{\frac{1}{q}}$, then $\displaystyle \displaystyle \lim_{x\to{a}}\dfrac{x^n-a^n}{x-a} = \lim_{x\to{a}}\frac{x^{\frac{p}{q}}-a^{\frac{p}{q}}}{x-a} = \lim_{x\to{a}}\frac{(x^{\frac{1}{q}})^p-(a^{\frac{1}{q}})^p}{x-a} = \lim_{y\to{b}}\frac{y^p-b^p}{y^q-b^q}$ $\displaystyle = \displaystyle \lim_{y\to{b}}\frac{(y^p-b^p)/(y-b)}{(y^q-b^q)/(y-b)}$ and by (1) this is $\displaystyle \dfrac{pb^{p-1}}{qb^{q-1}} = \frac{p}{q}b^{p-q} = \left(\frac{p}{q}\right)(b^p)^{(1-\frac{q}{p})} = na^{n-1}$.
3. If $\displaystyle n\in\mathbb{Q}_{< 0}$, then let $\displaystyle n = -m$. We have $\displaystyle \displaystyle \lim_{x\to{a}}\dfrac{x^n-a^n}{x-a} = \lim_{x\to{a}}\dfrac{x^{-m}-a^{-m}}{x-a} = \lim_{x\to{a}}\dfrac{(a^m-x^m)/(x^ma^m)}{x-a} = \lim_{x\to{a}}\left(\dfrac{-1}{x^ma^m}\right)\left(\dfrac{x^{m}-a^{m}}{x-a}\right)$ and by (2) this is $\displaystyle \frac{-1}{a^{2m}}\cdot ma^{m-1} = -ma^{-m-1} = na^{n-1}$.

How do you deal with this when $\displaystyle n\not\in\mathbb{Q}$? Or maybe there is a way of doing it on one-go?

Let $\displaystyle r\in\mathbb{R}-\mathbb{Q}\Longrightarrow\,\exists \{p_n\}\subset \mathbb{Q}\,\,s.t.\,\,p_n\xrightarrow [n\to\infty]{} r$ , so we can put:

$\displaystyle \lim\limits_{x\to a}\frac{x^r-a^r}{x-a}=\lim\limits_{n\to\infty}\lim\limits_{x\to a}\frac{x^{p_n}-a^{p_n}}{x-a}$ , and then you know what to do with rational powers.

Now, why can we put the limits as above? Because of continuity of the exponential functions, but we can do it also without it by means of Dedekind cuts, supremums and stuff, though I'm afraid this is not so common a subject in nowadays first years in mathematics depts.

Tonio

3. Super!

We can do it also without it by means of Dedekind cuts, supremums and stuff, though I'm afraid this is not so common a subject in nowadays first years in mathematics depts.
I'm not too familiar with the 'super mums' (pun intended) and Dedekind cuts. A new (analysis) book that I
bought has entire sections on this stuff, which I have not yet learned -- they don't come off as easy nor
particularly intuitive, which might explain why they are not popular in first academic years, but they are
quite fascinating and I hope to attend a university where this is not the case. Oh, well... Thanks, Tonio.