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let $k,n$ be a natural numbers with $k>n$ and $f_m(x)$ be a function such that

$$f_m(0)=0 , f_m^{(p)}(0)\in R , \space \forall \space m\in\{1,2,...,k\} \space \wedge \space p\in\{1,2,...,n\}$$

then how can I prove that

$$ \lim_{x\to 0} \frac{d^n}{dx^n} \prod_{m=1}^k f_m(x)=0$$ I used special case of that in this Question for $f_m(x)=e^x-1$

Leibniz Rule show that to derivative all $f_m(x)$ at least one time it need to be $k\ge n$

So if $k<n$ then there exist $f_m^{(0)}(x)$ which made it zero.

but this prove is not fine So how can I prove it in better way ?

and How can we prove it without using Leibniz Rule ?

Faoler
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    We generally say $f_m$ is a sequence of functions, and only call $f_m$ "a function" if we write it as $f(m,x)$ rather than $f_m(x).$ – Thomas Andrews Aug 22 '24 at 19:08
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    Why are you taking the limit of the $n^\text{th}$ derivative as $x\to 0$ rather than just calculating the derivative at $x=0$? And why would you want to avoid using the Leibniz rule? – Ted Shifrin Aug 22 '24 at 19:10
  • @TedShifrin taking the limit at zero or the the function at $x=0$ isn't the same ?! and I don't avoid that rule but its better to prove it without any theorem (without prove it) since the problem seems simple – Faoler Aug 22 '24 at 19:42
  • Maybe consider the Taylor polynomial of degree $n$ of the product function. – Ted Shifrin Aug 22 '24 at 23:16
  • @TedShifrin yes that's good idea I didn't think about that – Faoler Aug 23 '24 at 07:48

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