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?

  • 1
    $\begingroup$ You are missing a $-$ sign on the second line after 'Then'. $\endgroup$
    – copper.hat
    Apr 4 '18 at 16:44
  • 4
    $\begingroup$ Note that $\lim_{x \to x_0} \frac{x_0^n}{x-x_0} = \infty$ so you cannot use the product rule of limits in the way that you have. $\endgroup$
    – angryavian
    Apr 4 '18 at 16:50
  • $\begingroup$ I was going to point out as angryavian that your step from the line he mentions to the line where a limint that has that part disappeared magically doesn't make sense plus that limit diverges. $\endgroup$ Apr 4 '18 at 16:53
  • $\begingroup$ use this formula a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+....+b^{n-1}). $\endgroup$
    – Bernstein
    Apr 4 '18 at 16:53
  • $\begingroup$ You can not evaluate limits for parts of the expression, sum the rest and evaluate the sum as you have above. There are restrictions on when the conventional rules of algebra still apply inside the limit. $\endgroup$
    – Doug M
    Apr 4 '18 at 17:04

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.



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}


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