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$\underset{x→1}{\lim}(x-1)\{x\}$, 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 |
