solve-with-explanation-lim-x-0-x-sinx-where-represents-greatest-integer-

Question Number 74870 by vishalbhardwaj last updated on 02/Dec/19

solve with explanation

li \underset{x \to 0^{-}}{m} [\frac{x}{sinx}], where [] represents greatest integer

Commented by mathmax by abdo last updated on 02/Dec/19

w e h a v e s i n x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{(2 n + 1)!} w i t h r a d i u s R = + \infty

s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots . \Rightarrow x - \frac{x^{3}}{6} ⩽ s i n x ⩽ x (w e c a n t a k e 0 ⩽ x ⩽ \frac{π}{2})

1 - \frac{x^{2}}{6} ⩽ \frac{s i n x}{x} ⩽ 1 \Rightarrow [1 - \frac{x^{2}}{6}] ⩽ [\frac{s i n x}{x}] ⩽ 1 \Rightarrow 1 + [- \frac{x^{2}}{2}] ⩽ [\frac{s i n x}{x}] ⩽ 1

\Rightarrow {l i m}_{x \to 0} [\frac{s i n x}{x}] = 1

Answered by mind is power last updated on 02/Dec/19

\frac{x}{\sin (x)} > 0 whe x \in] - \frac{π}{2}, 0 [

[\frac{x}{\sin (x)}] ⩽ \frac{x}{\sin (x)} \dots \sqrt{E}

x < \sin (x), for all x \in] - \frac{π}{2}, 0 [

proof

\cos (t) ⩽ 1

f (x) = x - \sin (x) \Rightarrow f^{'} (x) = 1 - \cos (x) ⩾ 0

f increase f (0) = 0 \Rightarrow x - \sin (x) < 0 \forall x \in [- \frac{π}{2}, 0 [

\Rightarrow x < \sin (x)

\Rightarrow \frac{x}{\sin (x)} ⩾ 1, ∴ \sin (x) ⩽ 0 ∴

\Rightarrow [\frac{x}{\sin (x)}] ⩾ 1

\Rightarrow E \Leftrightarrow 1 ⩽ [\frac{x}{\sin (x)}] ⩽ \frac{x}{\sin (x)}

\lim_{x \to 0} \frac{x}{\sin (x)} = \lim_{x \to 0} \frac{1}{\cos (x)} = 1 hopitaks Rulls

\Rightarrow \lim_{x \to 0} [\frac{x}{\sin (x)}] = 1