Limits of tolerance

Calculus Level 3

lim x 0 ( 200 sin x x + 194 x tan x ) = ? { \lim_{x\to0} } \left( \left \lfloor 200 \cdot \frac { \sin { x } }{ x } \right \rfloor +\left\lfloor 194 \cdot \frac { x }{ \tan { x } } \right\rfloor \right) = \, ?

Notation : \lfloor \cdot \rfloor denotes the floor function .

400 0 392 6 394

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1 solution

Chew-Seong Cheong
Mar 22, 2018

L = lim x 0 ( 200 sin x x + 194 x tan x ) By Maclaurin series = lim x 0 200 x x 3 3 ! + x 5 5 ! x + lim x 0 194 x x + 1 3 x 3 + 2 15 x 5 + = lim x 0 ± 200 x x 3 3 ! + x 5 5 ! x + lim x 0 ± 194 x x + 1 3 x 3 + 2 15 x 5 + For both lim x 0 + and lim x 0 = 200 1 + lim x 0 ± 194 1 = 199 + 193 = 392 \begin{aligned} L & = \lim_{x \to 0} \left(\left \lfloor 200\cdot \frac {\color{#3D99F6}\sin x}x \right \rfloor + \left \lfloor 194\cdot \frac x{\color{#3D99F6}\tan x} \right \rfloor \right) & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to 0} \left \lfloor 200\cdot \frac {\color{#3D99F6}x-\frac {x^3}{3!}+\frac {x^5}{5!}-\cdots}x \right \rfloor +\lim_{x \to 0} \left \lfloor 194\cdot \frac x{\color{#3D99F6}x+ \frac 13x^3+\frac 2{15}x^5+\cdots} \right \rfloor \\ & = \lim_{\color{#D61F06}x \to 0^\pm} \left \lfloor 200\cdot \frac {\color{#3D99F6}x-\frac {x^3}{3!}+\frac {x^5}{5!}-\cdots}x \right \rfloor +\lim_{\color{#D61F06}x \to 0^\pm} \left \lfloor 194\cdot \frac x{\color{#3D99F6}x+ \frac 13x^3+\frac 2{15}x^5+\cdots} \right \rfloor & \small \color{#D61F06} \text{For both }\lim_{x \to 0^+} \text{ and }\lim_{x \to 0^-} \\ & = \left \lfloor 200\cdot {\color{#3D99F6}1^-} \right \rfloor +\lim_{\color{#D61F06}x \to 0^\pm} \left \lfloor 194\cdot {\color{#3D99F6}1^-} \right \rfloor \\ & = 199+193 = \boxed{392} \end{aligned}

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