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Question Number 183162 by mnjuly1970 last updated on 21/Dec/22

	Answered by aleks041103 last updated on 23/Dec/22
	$sin(x)=x−(x^3 /6)+o(x^3 ) ⇒−((sin(x))/x)=−1+(x^2 /6)+o(x^2 )→−1^+ −(x/(sin(x)))→−1^− ⇒⌊−((sin(x))/x)⌋→−1 ⌊−(x/(sin(x)))⌋→−2 lim_(x→0^± ) ⌊sin(x)⌋+⌊−sin(x)⌋=⌊0^± ⌋+⌊0^∓ ⌋= = { (0),((−1)) :}+ { ((−1)),(0) :}=−1 ⇒lim_(x→0) ((⌊−((sin(x))/x)⌋+⌊−(x/(sin(x)))⌋)/(⌊sin(x)⌋+⌊−sin(x)⌋))=((−1+(−2))/(−1))=3 lim_(x→0) ((⌊−((sin(x))/x)⌋+⌊−(x/(sin(x)))⌋)/(⌊sin(x)⌋+⌊−sin(x)⌋)) = 3$
	$s i n (x) = x - \frac{x^{3}}{6} + o (x^{3})$ $\Rightarrow - \frac{s i n (x)}{x} = - 1 + \frac{x^{2}}{6} + o (x^{2}) \to - 1^{+}$ $- \frac{x}{s i n (x)} \to - 1^{-}$ $\Rightarrow ⌊ - \frac{s i n (x)}{x} ⌋ \to - 1$ $⌊ - \frac{x}{s i n (x)} ⌋ \to - 2$ $\underset{x \to 0^{\pm}}{l i m} ⌊ s i n (x) ⌋ + ⌊ - s i n (x) ⌋ = ⌊ 0^{\pm} ⌋ + ⌊ 0^{\mp} ⌋ =$ $= {\begin{cases} 0 \\ - 1 \end{cases} + {\begin{cases} - 1 \\ 0 \end{cases} = - 1$ $\Rightarrow \underset{x \to 0}{l i m} \frac{⌊ - \frac{s i n (x)}{x} ⌋ + ⌊ - \frac{x}{s i n (x)} ⌋}{⌊ s i n (x) ⌋ + ⌊ - s i n (x) ⌋} = \frac{- 1 + (- 2)}{- 1} = 3$ $\underset{x \to 0}{l i m} \frac{⌊ - \frac{s i n (x)}{x} ⌋ + ⌊ - \frac{x}{s i n (x)} ⌋}{⌊ s i n (x) ⌋ + ⌊ - s i n (x) ⌋} = 3$
	Commented by mnjuly1970 last updated on 29/Dec/22

	$b r a v o s i r$
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