# Proving existence of limit $h \to 0$ of Newton's difference quotient, for all $x$ of $f(x)=x^n$

I had a question for calculus that, given a function $$f: \mathbb{R} \to \mathbb{R}$$ defined by $$f(x) = x^3$$, to find Newton's difference quotient, which is $$\frac{\Delta f(x)}{\Delta (x)} = 3x^2+3xh + h^2$$ I was then asked to show that the limit $$\lim_{h \to 0} \frac{\Delta f(x)}{\Delta (x)}$$ exists for all $$x$$. Now, I'm assuming that the actual question is just a one-liner where I can just use intuition to eliminate the last two terms and be done with it. But, I wanted to prove it using the epsilon-delta definition of limits - I did not get anywhere.

The popular resources online seem to show how to prove stuff with a single variable and use neat tricks like setting $$\delta$$ to the $$\min$$ of some values as a way to clamp it. But any time I apply those methods to this question, the fact that there is $$x$$ breaks it, because it can be negative or positive.

So, since I am asking anyway, might as well generalise the question to $$f(x)=x^n \ (n \in \mathbb{N}^+)$$. How so I prove that Newton's difference quotient of that function has a limit $$h \to 0$$ for all $$x$$?

The easiest way is to solve it term by term. Start by showing that $$\lim_\limits{h \to 0}3xh=0$$. If $$x=0$$ we have nothing to show, so suppose $$x \ne 0$$ ($$x$$ is arbitrary, but fixed). For any $$\epsilon>0$$ choose $$\delta_1 \le \frac{\epsilon}{3|x|}$$. For the second monomial choose $$\delta_2 \le \sqrt{\epsilon}$$ and take $$\delta:= \frac{1}{2}\min\{ \delta_1, \delta_2\}$$. The sign of $$x$$ doesn't matter, as you will always have that $$\forall h \in \left(-\delta_1, \delta_1\right) \left|3xh\right|< \epsilon$$ For the general case. If $$n \in \mathbb{N}$$ you have $$\frac{\Delta f(x)}{h}= \sum_{k=0}^{n-1}{n \choose k}h^{n-k-1}x^k$$ So, for $$k=0, \cdots , n-2$$ you can define $$\delta_k \le \left( {n \choose k}|x|^k \right)^{-\frac{1}{n-k-1}} \epsilon^{\frac{1}{n-k-1}}$$ and $$\delta:= \frac{1}{n-1}\min_k\{\delta_k\}$$
• This is unnecessarily complicated. If you suppose $|h|\leq 1$, then $|h|^2\leq |h|$, which gives a simpler bound, same thing for higher order polynomials. Dec 4, 2023 at 3:07