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If \mathrm{f(x)=\left\{\begin{array}{l} \frac{\sin \{x\}}{\{x\}},\{x\} \neq 0 \\ k, \quad\{x\}=0, \end{array}\right.} where x denotes fractional part of x , then f(x) will be continuous

Option: 1

\mathrm{\text { if } k=0}


Option: 2

\mathrm{\text { if } k=\sin 1}


Option: 3

\mathrm{\text { if } k=1}


Option: 4

\mathrm{\text { for no value of } k}


Answers (1)

best_answer

\mathrm{ \lim _{x \rightarrow 1+} f(x)=\lim _{y \rightarrow 0+} \frac{\sin y}{y}=1 }
and
\mathrm{ \lim _{x \rightarrow 1-} f(x)=\lim _{y \rightarrow 1-} \frac{\sin y}{y}=\frac{\sin 1}{1} \neq 1 }

Hence given limit does not exist for any value of k.

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Gaurav

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