Evaluate-the-following-limit-if-it-exists-lim-x-0-sin-sin-sinx-x-

Question Number 1817 by 112358 last updated on 05/Oct/15

E v a l u a t e t h e f o l l o w i n g l i m i t,

i f i t e x i s t s,

\lim_{x \to 0} \frac{s i n (s i n (s i n x))}{x} .

Answered by 123456 last updated on 05/Oct/15

f_{n} (x) = \sin f_{n - 1} (x)

f_{0} (x) = x

theorem : for n \in N

\lim_{x \to 0} \frac{f_{n} (x)}{x} = 1

proof :

\lim_{x \to 0} \frac{f_{0} (x)}{x} = \lim_{x \to 0} \frac{x}{x} = 1 (n = 0)

\lim_{x \to 0} \frac{f_{1} (x)}{x} = \lim_{x \to 0} \frac{\sin x}{x} = 1 (n = 1)

suppose that it is truth for n, them

\lim_{x \to 0} \frac{f_{n + 1} (x)}{x} = \lim_{x \to 0} \frac{\sin f_{n} (x)}{x}

= \lim_{x \to 0} \frac{\sin f_{n} (x)}{f_{n} (x)} \cdot \frac{f_{n} (x)}{x}

= \lim_{x \to 0} \frac{\sin f_{n} (x)}{f_{n} (x)} \lim_{x \to 0} \frac{f_{n} (x)}{x}

= \lim_{x \to 0} \frac{\sin f_{n} (x)}{f_{n} (x)}

y = f_{n} (x) = \sin \dots . \sin x

x \to 0 \equiv y \to 0

\lim_{x \to 0} \frac{\sin f_{n} (x)}{f_{n} (x)} = \lim_{y \to 0} \frac{\sin y}{y} = 1

cololary : \lim_{x \to 0} \frac{\sin \sin \sin x}{x}

proof :

\sin \sin \sin x = f_{3} (x)

\lim_{x \to 0} \frac{\sin \sin \sin x}{x} = \lim_{x \to 0} \frac{f_{3} (x)}{x} = 1 (by theorem)

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