limx→0([100xsinx]+[99sinxx])= (Where [.] represents the greatest integral function)

0 0;xsin⁡x>1 and ∀x 1 and sin⁡xx 100 and 99sin⁡xx<99100xsin⁡x=100,99sin⁡xx=98

limx→0([100xsinx]+[99sinxx])= (Where [.] represents the greatest integral function)

0 0;xsin⁡x>1 and ∀x 1 and sin⁡xx 100 and 99sin⁡xx<99100xsin⁡x=100,99sin⁡xx=98

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$\lim_{x \to 0} ([\frac{100 x}{\sin x}] + [\frac{99 \sin x}{x}]) =$ (Where [.] represents the greatest integral function)

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Detailed Solution

$0 < x < \frac{π}{2} \sin x < x < \tan x$

$\begin{array}{l} \forall x > 0; \frac{x}{\sin x} > 1 and \forall x < 0; \frac{\sin x}{x} < 1 \\ \frac{x}{\sin x} > 1 and \frac{\sin x}{x} < 1 \\ \Rightarrow \frac{100 x}{\sin x} > 100 and \frac{99 \sin x}{x} < 99 \\ [\frac{100 x}{\sin x}] = 100, [\frac{99 \sin x}{x}] = 98 \end{array}$