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  • If  <img alt="\mathrm{f(x)=\frac{\sin [x]}{[x]},[x] \neq 0,=0,[x]=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5Cfrac%7B%5Csin%20%5Bx%5D%7D%7B%5Bx%5D%7D%2C%

If  \mathrm{f(x)=\frac{\sin [x]}{[x]},[x] \neq 0,=0,[x]=0}. Where \mathrm{[x]}denotes the greatest integer less than or equal to \mathrm{x}, then \mathrm{ \lim _{x \rightarrow 0} f(x)} is equal to
 

Option: 1

\sin 1
 


Option: 2

0


Option: 3

\text{doesn't exist}
 


Option: 4

\text{none of these}


Answers (1)

best_answer

First note that by def. of \mathrm{[x]}, we have

\mathrm{[x]=-1 \: when \: -1 \leq x<0}

\mathrm{and \: [x]=0 \: when \: 0 \leq x<1.}

Hence by def. of \mathrm{f}, the function is re-defined as under :

\mathrm{f(x)=\frac{\sin (-1)}{-1}=\sin 1\: when -1 \leq x<0 \: and \: f(x)=0 \: \: \: \: when \: \: 0 \leq x<1.}

Change point is 0 and so we consider limit when \mathrm{x \rightarrow 0}

\mathrm{\therefore f(0-0)=\lim _{h \rightarrow 0} \sin 1=\sin 1.}

\mathrm{and \: f(0+0)=\lim _{h \rightarrow 0} 0=0.}

Since, \mathrm{f(0-0) \neq f(0+0)}, the limit of \mathrm{f(x)\: at \: x=0}does not exist.

Hence option 3 is correct.

Posted by

avinash.dongre

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