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\lim _{x \rightarrow 1} \frac{x \sin \{x\}}{x-1}, where {x} denotes the fractional part of x, is equal to 

Option: 1

-1


Option: 2

0


Option: 3

1


Option: 4

does not exist


Answers (1)

best_answer

\lim _{x \rightarrow 1-0}\{x\}=\lim _{x \rightarrow 1-0}(x-[x])=1-0=1 \\

\lim _{x \rightarrow 1+0}\{x\}=\lim _{x \rightarrow 1+0}(x-[x])=1-1=0 \\

\therefore \lim _{x \rightarrow 1-0} \frac{x \sin \{x\}}{x-1}=\lim _{x \rightarrow 1-0} \frac{x}{x-1} \sin \{x\}=-\infty \cdot \sin (1)=-\infty \\

Since L.H. limit \neq R.H. limit

\therefore limit does not exist.

Hence (d) is the correct answer.

 

Posted by

Sayak

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