Epsilon-delta derivative proof of $x^n$

I'm currently trying to prove the power rule using the epsilon-delta definition of a derivative. I've already done it for the basic limit definition, but I thought it might be a helpful exercise to test my understanding by doing it this way. However, I'm struggling and would really appreciate any help/hints.

My work so far:

The epsilon-delta definition says a function is differentiable if, for every $\epsilon > 0$, there exists a $\delta > 0$ such that $|x-x_0| < \delta$ implies $\frac{f(x)-f(x_0)}{x-x_0} - f'(x_0) < \epsilon$. Thus, I need to find a delta, as function of epsilon, and possibly $x_0$, such that whenever the first inequality is true the second one is as well.

So, fix $x_o \in R$, and let $|x-x_0| < \delta$, then $|\frac{x^n - x_{0}^{n}}{x-x_0} - nx_{0}^{n-1}| = |\frac{x^n - nxx_{0}^{n-1} + (n-1)x_{0}^{n}}{x-x_0}|$. Upper-bounding this with $x < \delta + x_0$ and simplifying, we get $|\frac{(\delta + x_0)^n - n\delta x_{0}^{n-1}-x_{0}^{n}}{\delta}| = |\frac{(\delta + x_0)^n}{\delta} -nx_{0}^{n-1} - x_{0}^{n}| < \epsilon$.

Is that right so far, and if so, any advice on how to push through this last step and solve for $\delta$? I'm stumped, to be honest.

• How denominator changed with $\delta$ to have upper bound? – Atbey Nov 22 '17 at 6:31
• Do a binomial expansion on $(\delta - x_0)^n$ the first two terms cancel with what is there, leaving $\delta |{n\choose 2} x_0^{n-2} +{n\choose 3} x_0^{n-3}\delta + \cdots + \delta^{n-2}| < \epsilon$ and show that everything inside the absolute value is bounded. – Doug M Nov 22 '17 at 6:58
• Your last line is incorrect. The middle term should be $|\frac {(\delta +x_0)^n -x_0^n}{\delta}-nx_0^{n-1}|$ – DanielWainfleet Nov 22 '17 at 10:53

Using

$$(x^n - x_0^n) = (x - x_0)\sum_{k=0}^{n-1} x^k x_0^{n-1-k}$$

One can write

$$\frac{x^n - x_0^n}{x - x_0} - n x_0^{n-1} = \sum_{k=0}^{n-1} (x^k x_0^{n-1-k} - x_0^{n-1})$$ Define $M = |x_0| + 1$ and suppose that $|x-x_0|\le 1$, then $|x|\le M$ and $|x_0|\le M$ and $$|x^k-x_0^k| \le |x - x_0|\sum_{p=0}^{k-1} |x|^p |x_0|^{k-1-p}\le |x - x_0|\sum_{p=0}^{k-1} M^{k-1}\le |x-x_0|k M^{k-1}$$

We get $$\left|\frac{x^n - x_0^n}{x - x_0} - n x_0^{n-1}\right| \le \sum_{k=1}^{n-1} |x-x_0|k M^{k-1} |x_0|^{n-1-k}\le |x - x_0|M^{n-2}\sum_{k=1}^{n-1}k \le |x - x_0|M^{n-2}\frac{n (n-1)}{2}$$

For $n>1$ we can take $$\delta = \min\left(1, \frac{2\varepsilon}{n(n-1)(|x_0|+1)^{n-2}}\right)$$ For $n=1$ there is no condition on $\delta$ because $f(x) = f(x_0) + (x-x_0)$.

• Using this hint at the beginning, I get $|\frac{x^n - x_{0}^{n}}{x-x_0} - nx_{0}^{n-1}| = |(1-n)x_{0}^{n-1} + xx_{0}^{n-2} + ... + x_0x^{n-2} + x^{n-1}| < \epsilon$. Is this what you had in mind? I don't understand how this helps, since there doesn't seem to be an easy way to cancel all of the x terms floating around. – P. Reinecke Nov 22 '17 at 6:57
• See my edit above. – Gribouillis Nov 22 '17 at 7:15

For $x\ne 0$ we have $$(x+d)^n-x^n=-x^n+\sum_{j=0}^n x^{n-j}d^j\binom {n}{j}=nx^{n-1}d +d^2 P(d)$$ where $P(d)$ is a polynomial. So there exists $M\in \Bbb R^+$ such that $\forall d\; (|d|\leq 1\implies |P(d)|<M).$ Therefore for $0<|d|\leq 1$ we have $$|\frac {(x+d)^n-x^n}{d}-nx^{n-1}| =$$ $$= |\frac {nx^{n-1}d+d^2P(d)}{d}-nx^{n=1}| =$$ $$= |dP(x)|\leq |d|M.$$ And $|d|M\to 0$ as $d\to 0.$

To prove that $M$ exists : If $P(d)=\sum_{j=0}^ma_jd^j$ then $|d|\leq 1\implies |P(d)|\leq$ $\sum_{j=0}^m|a_j|\cdot |d|^j\leq$ $\leq \sum_{j=0}^m|a_j|=M.$

For $x=0\ne d$ we have $\frac {(x+d)^n-x^n}{d}=\frac {(0+d)^n-0^n}{d}=d^{n-1}.$

• Note that the degree of $P(d)$ and the values of the co-efficients of $P(d)$ are determined by $x$ and $n$ but are independent of $d.$ – DanielWainfleet Nov 22 '17 at 11:40