$\underset{x→1}{\lim}\frac{x\sin\{x\}}{x-1}$, where {x} denotes the fractional part of x, is equal to

does not exist

for $\{x\}$

$LHL=\underset{x→1^-}{\lim}\{x\}=\underset{x→1^-}{\lim}(x-[x])=1-0=+1$

$RHL=\underset{x→1^+}{\lim}\{x\}=\underset{x→1^+}{\lim}(x-[x])=1-1=0$

for function

$LHL=\underset{x→1^-}{\lim}\frac{x\sin\{x\}}{x-1}=\frac{\sin\{x\}}{0^-}=-∞$

$RHL=\underset{x→1^-}{\lim}\frac{x\sin\{x\}}{x-1}=\frac{x\sin\{x\}}{\{x\}}=1$

LHL ≠ RHL ⇒ limit doesn't exist