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Let's consider $f, g: (0, +\infty) \rightarrow\mathbb{R}$, $f(x)=\displaystyle\frac{\sin x}{x}$, $g(x)=\displaystyle\frac{\cos x}{x}$. Find the following limits

$$\lim_{n\to\infty}f^{(n)}(x)$$ $$\lim_{n\to\infty}g^{(n)}(x)$$

where $f^{(n)}$ and $g^{(n)}$ are the $n$th derivatives of $f(x)$, respectively $g(x)$.
It's a problem I thought of last days and I didn't guess the answer by trying to look at the first derivatives of both functions. What should I do here to get the limits? Thanks.

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  • $\begingroup$ Do you assume this has a closed form ? $\endgroup$ – mick Sep 29 '12 at 18:26
  • $\begingroup$ You can easily give a boundary by using taylor's theorem , but i guess you knew that already. $\endgroup$ – mick Sep 29 '12 at 18:28
  • $\begingroup$ Maybe this is naive but if you use hypergeo form I think you can solve this. $\endgroup$ – mick Sep 29 '12 at 18:33
  • $\begingroup$ As you take derivatives, the power of $x$ in the denominator will grow without bound. $\endgroup$ – Graphth Sep 29 '12 at 18:35
  • $\begingroup$ Try exploiting $xy=\sin x$ and differentiate $n$ times using Leibniz's rule. $\endgroup$ – Pedro Tamaroff Sep 29 '12 at 18:38
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$f'(x) = -x^{-2}\sin x + x^{-1}\cos x$ suggests that $f^{(n)}(x)$ can be written as $P_n(x^{-1})\sin x + Q_n(x^{-1})\cos(x)$ with polynomials $P_n, Q_n\in \mathbb Z[X]$. Indeed, the derivative of $P_n(x^{-1})\sin x + Q_n(x^{-1})\cos(x)$ is $-x^{-2}P_n'(x^{-1})\sin x+P_n(x^{-1})\cos x -x^{-2}Q_n'(x^{-1})\cos(x)-Q_n(x^{-2})\sin x$, so that we are led to the recursions $$ P_{n+1}=-X^2P_n'-Q_n,\\Q_{n+1}=P_n-X^2Q_n'.$$ Letting $R_n=i^{-n}(P_n+iQ_n)\in \mathbb Z[i,X]$, we see that $$\tag1 R_{n+1}=iX^2R_n'+R_n.$$ By induction one readily shows (starting with $R_0=X$)

$$ R_n = \sum_{k=0}^ni^k\frac{n!}{(n-k)!}X^{k+1}.$$ While this allows us to write down $f^{(n)}(x)$ and $g^{(n)}(x)$ explicitly, there is no hint that $\lim_{n\to\infty}f^{(n)}(x)$ or $\lim_{n\to\infty}g^{(n)}(x)$ should exist for any $x$ (not even for multiples of $\frac\pi2$).

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Consider the function $$h(x):={e^{ix}\over x}\ .$$ Computing the first few derivatives using paper and pencil one is lead to the conjecture that $$h^{(n)}(x)={1\over x}p_n\Bigl({1\over x}\Bigr)\ e^{ix}\ ,$$ where $p_n(t)=\sum_{k=0}^n c_k\ t^k$ is a polynomial of degree $\leq n$ with complex coefficients.

This conjecture can be proven by induction, and then the original claim about $f$ and $g$ is immediate.

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