Proof verification: derivative of $x^n$ Let $X\subset\mathbb{R}$, $n\in \mathbb{N}$ and $f:X\to \mathbb{R}, x \mapsto x^n$. I have to show that $f'(x) = n \cdot x^{n-1}$. This follows easily by combining induction and product rule for derivatives. However I have yet another proof:
(induction step)
Suppose $\lim_{x\to x_0} \frac{x^{n} - x_{0}^{n}}{x - x_0} = n \cdot x^{n-1}$.
Then \begin{equation} \begin{array}{ll} 
\lim_{x\to x_0} \frac{x^{n+1} - x_{0}^{n+1}}{x - x_0} &= \lim_{x\to x_0} \frac{x^{n}\cdot x - x_{0}^{n} \cdot x_0}{x - x_0}\\
&=\lim_{x\to x_0} \frac{x^{n}}{x - x_0} \cdot x + \frac{x_{0}^{n}}{x-x_0} \cdot x_0\\
&=\lim_{x\to x_0} \frac{x^{n}}{x - x_0} \cdot \lim_{x\to x_0} x + \lim_{x\to x_0} \frac{x_{0}^{n}}{x-x_0} \cdot x_0x\\
&= x_0 \cdot \lim_{x\to x_0} \frac{x^{n}}{x - x_0} + x_0 \cdot \lim_{x\to x_0}
\frac{x_{0}^{n}}{x-x_0}\\
&= x_0 \cdot \left(\lim_{x\to x_0} \frac{x^{n}}{x - x_0} + \lim_{x\to x_0} \right)\\
&= x_0 \cdot \lim_{x\to x_0} \frac{x^{n} - x_{0}^{n}}{x - x_0}\\
&= x_0 \cdot n \cdot x_0^{n-1}\\
&= n \cdot x_0^{n}
\end{array} \end{equation}
Which is obviously false. Where does the mistake(s) lurk?
 A: You can do this:
Assume
$\lim_\limits{x\to x_0} \frac{x^{n} - x_{0}^{n}}{x - x_0} = nx_0^{n-1}$
$\lim_\limits{x\to x_0} \frac{x^{n+1} - x_{0}^{n+1}}{x - x_0} = $$\lim_{x\to x_0} \frac{x^{n}\cdot x - x_{0}^{n} \cdot x_0}{x - x_0}\\
\lim_\limits{x\to x_0} \frac{x^{n}\cdot x -x_{0}^{n} \cdot x+x_{0}^{n} \cdot x- x_{0}^{n} \cdot x_0}{x - x_0}\\
\lim_\limits{x\to x_0} \frac{(x^{n} -x_{0}^{n})x +x_{0}^{n}(x-x_0)}{x - x_0}\\
\lim_\limits{x\to x_0} (\frac{(x^{n} -x_{0}^{n})}{x-x_0})x +\lim_\limits{x\to x_0}\frac {x_{0}^{n}(x-x_0)}{x - x_0}\\
(nx_0^{n-1})(x_0) + x_0^n$
Applying the inductive hypothesis to the last step.
$(n+1)x_0^{n}$
A: You can only split a limit into a product of two limits if both respective limits exist.
instead, using continuity of polynomials the proof for your induction step becomes simple(although this does not require the induction hypothesis, which hints that induction is unnecessary):
\begin{equation} \begin{array}{ll} 
\lim_{x\to x_0} \frac{x^{n+1} - x_{0}^{n+1}}{x - x_0} &= \lim_{x\to x_0} \frac{(x-x_0)(x^n+x^{n-1}x_0+....+x_0^n)}{x - x_0}\\
&= \lim_{x\to x_0} (x^n+x^{n-1}x_0+....+x_0^n)\\
&= (n+1) \cdot x_0^{n}
\end{array} \end{equation}
