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Compute $\sin(\sin(...\sin(x)))$

Problem Find the limit $\sin(\sin(...\sin(x)))$ for any real number $x$. Solution The problem can be precisely formulated as follows: Find the limit $L$ such that $$ L \equiv\lim_{n\rightarrow\infty}{f_n(x)}, $$ where $x\in\mathbb{R}$, $n\in\mathbb{N}$, and $$\begin{align} f_1(x) &= \sin(x)\\ f_n(x) &= \sin(f_{n-1}(x)). \end{align}$$ We observe that, if the limit $L$ exists, $$\begin{align} \lim_{n\rightarrow\infty} f_{n-1}(x) &= L\\ \lim_{n\rightarrow\infty} f_n(x) &= L, \end{align}$$ and thus, $$ L = \sin{L}, $$ which we show to be $L=0$ in the following. First, we show that the limit exists. We accept the Axiom of Continuity of the Real Line, which says that any bounded monotonic sequence converges. Thus, to show that $L$ exists, we need only show that (1) $f_n(x)$ is a bounded sequence of $n$, and that (2) $f_n(x)$ is a monotonic sequence of $n$. $f_n(x)$ is clearly bounded because $f_n(...
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Find the limit of the sequence. $a_1=1, a_{n+1} = \frac{3a_n+4}{2a_n+3}$

We take the limits of sequences defined recursively by first showing the existence of the limits then, actually computing them. Existence is demonstrated by showing that the sequence is both (1) monotonic and (2) bounded. Problem Find the limit of the following sequence. $$\begin{align} $a_1&=1\\ a_{n+1} &= \frac{3a_n+4}{2a_n+3}$ \end{align}$$ Solution Monotonicity : We use mathematical induction to show that for all $n\in\mathbb{N}$, $a_{n+1} - a_n \geq 0$ and therefore, $\{a_n\}$ is monotonic increasing. For the base case at $n=1$, it is easy to see that $a_{n+1}-a_n = \frac{7}{5} - 1 = \frac{2}{5} >0.$ For the inductive case at $n=k$, we assume that $$\begin{align} a_{k+1} - a_{k} & = \frac{3a_k+4}{2a_k+3} - \frac{3a_{k-1}+4}{2a_{k-1}+3}\\ & = \frac{a_k - a_{k-1}}{(2a_k+3)(2a_{k-1}+3)} \geq 0. \end{align}$$ For $n=k...
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Prove: $\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}$, $\arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}}$

We prove identities involving inverse trigonometric functions. Sum formulas from trigonometry are exploited. $\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}$ Suppose that $$ \arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} \neq \frac{\pi}{4}. $$ Then, $$\begin{align} \tan{\left(\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}}\right)} &\neq \tan{\frac{\pi}{4}}\\ \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}} &\neq 1 \\ \frac{5}{6-5} &\neq 1 \\ 1 &\neq 1, \end{align}$$ which is a contradiction. Thus, $$ \arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}.□ $$ $\arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}}$ Suppose that $$ \arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} \neq \arcsin{\frac{3}{5}}. $$ Then, $$\begin{align} \sin{\left(\arcsin{\frac{16}{65}}+\arcsin{\fra...
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