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The following is a problem proposed in Pólya and Szegő's book "Problems and Theorems in Analysis"

Assume that $0<f(x)<x$ and $$f(x)=x-ax^k+bx^\ell+x^\ell \varepsilon(x),\,\;\;\;\lim_{x\to 0}\varepsilon(x)=0$$ for $0<x<x_0$, where $1<k<\ell$ and $a,b$ positive. Put $$v_0=x,\;\; v_1=f(v_0),\;\;v_2=f(v_1),\ldots,\;\;\;v_n=f(v_{n-1}),\ldots$$ Then we have for $n\to\infty$$$n^{\frac{1}{k-1}}v_n\to[(k-1)a]^{\frac{-1}{k-1}}$$

PROOF I will prove the better looking$$nv_n^{k-1}\to \frac{1}{a(k-1)}$$

To this end, first note that $v_n\to 0$, since $v_n$ decreases because of $0<f(x)<x$ and because of continuity at the origin, along with $f(0)=0$. Also note $f(x)/x\to 1$. Now, we have after some algebraic meddling that $$\frac{1}{{{x^{k - 1}}}} - \frac{1}{{f{{\left( x \right)}^{k - 1}}}} = \frac{{f{{\left( x \right)}^{k - 1}} - {x^{k - 1}}}}{{{x^{k - 1}}f{{\left( x \right)}^{k - 1}}}} = \frac{{f\left( x \right) - x}}{{{x^k}}}\frac{{f\left( x \right)}}{x}{\sum\limits_{m = 0}^{k - 2} {\left( {\frac{x}{{f\left( x \right)}}} \right)} ^m}$$

It follows that $$\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{f{{\left( x \right)}^{k - 1}}}} - \frac{1}{{{x^{k - 1}}}}} \right) = a\left( {k - 1} \right)$$

Thus, since $v_n\to 0$, we must have $$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{v_{n + 1}^{k - 1}}} - \frac{1}{{v_n^{k - 1}}}} \right) = a\left( {k - 1} \right)$$

Appealing to Cesàro's theorem, it follows that $$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\sum\limits_{m = 0}^{n - 1} {\frac{1}{{v_{m + 1}^{k - 1}}} - \frac{1}{{v_m^{k - 1}}}} } \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\frac{1}{{v_n^{k - 1}}} - \frac{1}{{v_0^{k - 1}}}} \right) = a\left( {k - 1} \right)$$

Which is what we wanted to prove. $\blacktriangle$

DSC The above is inspired in this particular proof when $f(x)=\sin x$.

Do you know any other proofs?

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I guess you are aware that Pólya and Szegő provide full solutions to almost all of the exercises. – Martin Jun 23 '13 at 1:20
@Martin Yep, I "have" the book. But the proof there looked quite strange, I didn't like it much. – Pedro Tamaroff Jun 23 '13 at 1:23
@Peter : Did you understand it or did you just not like it? Sometimes I don't like proofs because I feel like the ideas are not natural ; once I understand why they are natural in another context I like them better. – Patrick Da Silva Jun 23 '13 at 2:14
@PatrickDaSilva Fair enough, would you like to discuss their proof in the chat? – Pedro Tamaroff Jun 23 '13 at 2:27
@PatrickDaSilva Now? =P – Pedro Tamaroff Jun 23 '13 at 19:39

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