1
$\begingroup$

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 $x$?

$\endgroup$
0

1 Answer 1

0
$\begingroup$

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 $\delta:= \frac{1}{n-1}\min_k\{\delta_k\}$

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – peek-a-boo
    Dec 4, 2023 at 3:07
  • $\begingroup$ @peek-a-boo true, but this methods Is easily generalizzabile to sum of generic functions, not only monomials. Moreover It is the most "natural" bound I could think of. $\endgroup$
    – Marco
    Dec 4, 2023 at 3:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .