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Question Number 195126 by Erico last updated on 25/Jul/23

Soit f_{n} (x) = 2^{n + 1} [\frac{\frac{1}{2^{n}} c o t a n (\frac{x}{2^{n}}) - c o t a n x}{s i n (\frac{x}{2^{n}})}]

Double subscripts: use braces to clarify

Answered by MM42 last updated on 25/Jul/23

f_{n} (x) = \frac{2 s i n x c o s (\frac{x}{2^{n}}) - 2^{n + 1} s i n (\frac{x}{2^{n}}) c o s x}{{s i n}^{2} (\frac{x}{2^{n}}) s i n x}

= \frac{s i n (x + \frac{x}{2^{n}}) + s i n (x - \frac{x}{2^{n}}) - 2^{n} s i n (\frac{x}{2^{n}} + x) - 2^{n} s i n (\frac{x}{2^{n}} - x)}{{s i n}^{2} (\frac{x}{2^{n}}) s i n x}

= \frac{(1 - 2^{n}) s i n (\frac{1 + 2^{n}}{2^{n}}) x - (1 + 2^{n}) s i n (\frac{1 - 2^{n}}{2^{n}})}{{s i n}^{2} (\frac{x}{2^{n}}) s i n x}

= 2^{n} \times \frac{\frac{1 - 2^{n}}{2^{n}} s i n (\frac{1 + 2^{n}}{2^{n}}) x - \frac{1 + 2^{n}}{2^{n}} s i n (\frac{1 - 2^{n}}{2^{n}})}{{s i n}^{2} (\frac{x}{2^{n}}) s i n x}

T i p : i f x \to 0 \Rightarrow a s i n b x - b s i n a x \sim \frac{a b (a - b) (a + b)}{6} x^{3}

\Rightarrow {l i m}_{x \to 0} f_{n} (x) = {l i m}_{x \to 0} 2^{n} \times \frac{(\frac{1 - 2^{n}}{2^{n}}) (\frac{1 + 2^{n}}{2^{n}}) (\frac{1 - 2^{n}}{2^{n}} - \frac{1 + 2^{n}}{2^{n}}) (\frac{1 - 2^{n}}{2^{n}} + \frac{1 + 2^{n}}{2^{n}}) \frac{x^{3}}{6}}{{(\frac{x}{2^{n}})}^{2} \times x}

= \frac{(1 - 2^{2 n}) (- 2^{n + 1}) \times 2^{3 n} (2)}{2^{4 n} \times 6} =

= \frac{2^{2 n + 1} - 2}{3} ✓

{l i m}_{n \to \infty} \frac{f_{n} (x)}{2^{2 n + 2}} = \frac{1}{6} ✓

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