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Question Number 201657 by LimPorly last updated on 10/Dec/23

i f f (x) = {\begin{cases} \frac{\sin (1 + [x])}{[x]} f o r [x] \neq 0 \\ 0 f o r [x] = 0 \end{cases}

w h e r e [x] r e p r e s e n t s a n i n t e g e r x g r e a t e s t ⩽ x

F i n d \lim_{x \to 0^{-}} f (x) .

Answered by aleks041103 last updated on 10/Dec/23

I f I u n d e r s t a n d c o r r e c t l y :

[x] = m a x {n : n \in Z \land x ⩾ n}

i . e .

[2] = 2; [3.5] = 3; [- 1.5] = - 2 a n d s o o n .

T h e n

\underset{x \to 0^{-}}{l i m} [x] = \underset{ε \to 0^{+}}{l i m} [0 - ε] = \underset{ε \to 0^{+}}{l i m} [- ε] = - 1

\Rightarrow \underset{x \to 0^{-}}{l i m} f (x) = {({\begin{cases} \frac{\sin (1 + [x])}{[x]} f o r [x] \neq 0 \\ 0 f o r [x] = 0 \end{cases})}_{[x] = - 1} =

= \frac{s i n (1 + (- 1))}{(- 1)} = \frac{s i n (0)}{- 1} = 0

Commented by LimPorly last updated on 10/Dec/23

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