## Subject III. (20 points)

Let the function $f:(-1,1) \rightarrow R, f(t)=\frac{2 t}{1-t^{2}}$. Calculate:
a) $L_{1}=\lim _{x \rightarrow 0} \frac{1}{x^{n}} \prod_{k=1}^{n}(\underbrace{f \circ f \circ \ldots \circ f}_{\text {k times }})(\operatorname{tg} x)$;
b) $L_{2}=\lim _{x \rightarrow 0} \frac{1}{x} \sum_{k=1}^{n} \ln (1+(\underbrace{f \circ f \circ \ldots \circ f}_{\text {k times }})(\operatorname{tg} x))$, where $x \in\left(-\frac{\pi}{2^{n+1}}, \frac{\pi}{2^{n+1}}\right), n \in N^{*}$ fixed.

Prof. Ilie Diaconu, "Avram Iancu" Theoretical High School, Cluj-Napoca
## Subject III.

It is immediately shown, by induction, that: $\underbrace{\text { (fofo } \ldots \text { of })}_{\text {k times }}(\operatorname{tgx})=\operatorname{tg} 2^{k} x,(\forall) k \in N^{*}$.

Then we obtain:

a) $L_{1}=\lim _{x \rightarrow 0} \frac{1}{x^{n}} \prod_{k=1}^{n} \operatorname{tg}^{k} x=\lim _{x \rightarrow 0} \prod_{k=1}^{n} \frac{\operatorname{tg} 2^{k} x}{2^{k} x} \cdot \prod_{k=1}^{n} 2^{k}=2^{1+2+\cdots+n}=2^{\frac{n(n+1)}{2}}$;

b) $L_{2}=\lim _{x \rightarrow 0} \frac{1}{x} \sum_{k=1}^{n} \ln \left(1+\operatorname{tg} 2^{k} x\right)=\lim _{x \rightarrow 0} \sum_{k=1}^{n} \frac{\ln \left(1+\operatorname{tg} 2^{k} x\right)}{\operatorname{tg} 2^{k} x} \cdot \frac{\operatorname{tg} 2^{k} x}{2^{k} x} \cdot 2^{k}=\sum_{k=1}^{n} 2^{k}=2^{n+1}-2$.