Prove that $\lim_{\Delta x\to 0} \frac{\Delta ^{n}f(x)}{\Delta x^{n}} = f^{(n)}(x).$

Question

If $\Delta f(x)=f(x+\Delta x)-f(x)$, $(a)$ prove that $$\Delta\{\Delta f(x)\}=\Delta^2f(x)=f(x+2\Delta x)-2f(x+\Delta x)+f(x);$$ $(b)$ derive an expression for $\Delta^n f(x)$ where $n$ is any positive integer; and $(c)$ show that $$\lim\limits_{\Delta x\to0}\dfrac{\Delta^n f(x)}{(\Delta x)^n}=f^{(n)}(x)$$ if this limit exists.

I was able to prove $(a)$, and this is the expression I derived for $(b)$

$$\Delta ^{n}f(x)=\sum_{i=0}^{n}(-1)^{n-i}\binom{n}{i}f(x+i\Delta x)$$

I an fairly sure that the above is correct.

However, I am not sure how to prove,

$$\lim_{\Delta x\rightarrow 0} \frac{\Delta ^{n}f(x)}{\Delta x^{n}} = \lim_{\Delta x\rightarrow 0} \frac{\sum_{i=0}^{n}(-1)^{n-i}\binom{n}{i}f(x+i\Delta x)}{\Delta x^{n}} = f^{(n)}(x)$$

Answer 1

To elaborate on the my comment, I am posting an answer. Let's us take the case of $n = 2$ then we know that $$\Delta^{2} f(x) = f(x + 2\Delta x) - 2f(x + \Delta x) + f(x)$$ and hence

$\displaystyle \begin{aligned}\lim_{\Delta x \to 0}\frac{\Delta^{2}f(x)}{\Delta x^{2}} &= \lim_{\Delta x \to 0}\frac{f(x + 2\Delta x) - 2f(x + \Delta x) + f(x)}{\Delta x^{2}}\\ &\text{(apply L'Hospital Rule, differentiation wrt } \Delta x, x \text{ is constant)}\\ &= \lim_{\Delta x \to 0}\frac{2f'(x + 2\Delta x) - 2f'(x + \Delta x)}{2\Delta x}\\ &= \lim_{\Delta x \to 0}\frac{f'(x + 2\Delta x) - f'(x + \Delta x)}{\Delta x}\end{aligned}$

Now we need the assumption of existence of $f^{(n)}(x) = f''(x)$ here. We have $$ f''(x) = \lim_{\Delta x \to 0}\frac{f'(x + \Delta x) - f'(x)}{\Delta x}$$ hence we can see that $f'(x + \Delta x) = f'(x) + \Delta x\{f''(x) + \rho\}$ where $\rho$ denotes an expression which tends to $0$ with $\Delta x$. Replacing $\Delta x$ by $2\Delta x$ we see that $f'(x + 2\Delta x) = f'(x) + 2\Delta x\{f''(x) + \rho'\}$ where $\rho'$ is another expression which tends to $0$ with $\Delta x$. It now follows that $$f'(x + 2\Delta x) - f'(x + \Delta x) = \Delta x\{f''(x) + 2\rho' - \rho\}$$ and therefore $$\frac{f'(x + 2\Delta x) - f'(x + \Delta x)}{\Delta x} = f''(x) + 2\rho' - \rho$$ Taking limits when $\Delta x \to 0$ we get $$\lim_{\Delta x \to 0}\frac{f'(x + 2\Delta x) - f'(x + \Delta x)}{\Delta x} = f''(x)$$ and therefore we can see that $$\lim_{\Delta x \to 0}\frac{\Delta^{2}f(x)}{\Delta x^{2}} = f''(x)$$ The above proof can be carried in the same manner for any value of $n$ by repeated application of L'Hospital's rule, but at the last step when the denonominator $\Delta x^{n}$ reduces to $\Delta x$ we must make use of the existence of $f^{(n)}(x)$ the way explained above. In this connection please observe that we can't use L'Hospital rule at the last step to change $\Delta x $ in denominator to $1$ because L'Hospital Rule assumes that both numerator and denominator are differentiable in a neighborhood of the point under consideration (except possibly at that point). Hence we have to make use of the existence of $f^{(n)}(x)$ at a single point $x$ in the manner I have indicated above.

Note: The above solution would look better and more understandable if we replace $x$ by $a$ and $\Delta x$ by $h$, but I have tried to keep the notation aligned with that of OP.