$\underset{x→1}{\lim}(x-1)\{x\}$, where {.} denotes the fractional part, is equal to:

1 0 does not exist None of these

0

LHL = $\underset{h→0}{\lim}(1-h-1)\{1-h\}$ $=\underset{h→0}{\lim}(-h)\{1-h\}=0$ RHL = $\underset{h→0}{\lim}(1+h-1)\{1+h\}$ $=\underset{h→0}{\lim}h^2=0$ As LHL = RHL = 0, limiting value = 0