Let $\displaystyle f(x)=x^n$ and prove that its derivative is given by $\displaystyle f'(x)=n\cdot x^{n-1},\,\forall\,n\in\mathbb R.$
Hello,
Generalized Newton's binomial formula :
$\displaystyle (x+y)^n=\sum_{k=0}^\infty {n \choose k} x^k y^{n-k}$
where $\displaystyle {n \choose k}=\frac{r(r-1)\dots(r-k+1)}{k!}$ and $\displaystyle \forall n \in \mathbb{R}$
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A way of doing it is to remember another formula for the derivative :
(assuming it is differentiable so at least $\displaystyle x_0 \neq 0$)
$\displaystyle f'(x_0)=\lim_{x \to x_0} ~ \frac{x^n-x_0^n}{x-x_0}$
$\displaystyle \frac{x^n-x_0^n}{x-x_0}=\frac{x_0^n}{x_0} \times \frac{\left(\frac{x}{x_0}\right)^n-1}{\frac{x}{x_0}-1}=x_0^{n-1} \times \sum_{k=0}^{n-1} \left(\frac{x}{x_0}\right)^k$
$\displaystyle \lim_{x \to x_0} ~ x_0^{n-1} \times \sum_{k=0}^{n-1} \left(\frac{x}{x_0}\right)^k=x_0^{n-1} \times \sum_{k=0}^{n-1} 1^k=x_0^{n-1} \times n \quad \square$
This is my approach:
$\displaystyle \begin{aligned}
\frac{(x+\Delta x)^{n}-x^{n}}{\Delta x}&=\frac{e^{n\ln (x+\Delta x)}-e^{n\ln x}}{\Delta x} \\
& =x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{\Delta x} \\
& =x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}\cdot \frac{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}{\Delta x},
\end{aligned}$
hence,
$\displaystyle \frac{(x+\Delta x)^{n}-x^{n}}{\Delta x}=x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}\cdot \frac{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}{\dfrac{\Delta x}{x}}\cdot \frac{1}{x}\to nx^{n-1}$ as $\displaystyle \Delta x\to0.\quad\blacksquare$
Pardon me for that, but I do not find it elegant
First of all, why don't you directly factor by $\displaystyle x^n$ in the beginning instead of factorizing with exponentials ?
Also, what if $\displaystyle x+\Delta x < 0$ ?
Plus, you ought to know the limits $\displaystyle \lim_{x \to 0} ~ \frac{\ln(1+x)}{x}=1$ and $\displaystyle \lim_{x \to 0} ~ \frac{e^x-1}{x}=1$, which are proved using the definitions of derivative numbers (if not "advanced" calculus...)
Just giving my point.
You factorize later, after changing into exponentials, while it would have been exactly the same if you did it from the beginning.To use known limits.
$\displaystyle \ln(x+\Delta x)$ when $\displaystyle x + \Delta x< 0$ ?Why should I consider that case?
What's the interest of asking to prove a simple formula if it has to use nested limits ?By just knowing that $\displaystyle \lim_{x\to\infty}\left(1+\frac1x\right)^x=e$ those limits are easy to prove.
Well, I'm no one to tell you this, so just ignore it if you want.
Krazalid is correct to say limit definition is no good if $\displaystyle n$ is not an integer.
And generalized binomial theorem is faulty since you implicity use the power rule in its derivation.
The standard way is to notice that by definition $\displaystyle x^r = \exp(r\ln x)$ for $\displaystyle x>0$.
Here $\displaystyle \exp$ is the inverse function for $\displaystyle \ln t = \smallint_1^t \tfrac{du}{u}$.
If we are familar with the basic rules for $\displaystyle \ln$ and $\displaystyle \exp$ just apply the chain rule and get your answer.