$\underset{x→1}{\lim}(x-1)\{x\}$, where {.} denotes the fractional part, is equal to:
Answer & explanation
Correct answer: option 2
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