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Question: If \(f(x) = \left\{ \begin{aligned} & \begin{matrix} \frac{\sin\lbrack x\rbrack}{\lbrack x\rbrack},...

If f(x)={sin⁡[x][x],[x]≠00,[x]=0 f(x) = \left\{ \begin{aligned} & \begin{matrix} \frac{\sin\lbrack x\rbrack}{\lbrack x\rbrack}, & \lbrack x\rbrack \neq 0 \end{matrix} \\ & \begin{matrix} 0, & \lbrack x\rbrack = 0 \end{matrix} \end{aligned} \right.\ , then lim⁡x→0f(x)\lim_{x \rightarrow 0}f(x) equals

A

1

B

0

C

–1

D

Does not exist

Answer

Does not exist

Explanation

Solution

In closed interval of x=0x = 0 at right hand side [x] =0 and at left hand side [x]=−1.\lbrack x\rbrack = - 1. Also [0] =0.

Therefore function is defined as f(x)={sin⁡[x][x],(−1≤x<0)0,(0≤x<1) f(x) = \left\{ \begin{matrix} \frac{\sin\lbrack x\rbrack}{\lbrack x\rbrack}, & ( - 1 \leq x < 0) \\ 0, & (0 \leq x < 1) \end{matrix} \right.\

∴\therefore Left hand limit =lim⁡x→0−f(x)=lim⁡x→0−sin⁡[x][x]=sin⁡(−1)−1=sin⁡1c= \lim_{x \rightarrow 0 -}f(x) = \lim_{x \rightarrow 0 -}\frac{\sin\lbrack x\rbrack}{\lbrack x\rbrack} = \frac{\sin( - 1)}{- 1} = \sin 1^{c}Right hand limit = 0, Hence, limit doesn’t exist.