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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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    However i think that if we replace $n$ with factorial($n$) we do have a limit ?2012-09-29

2 Answers 2

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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.