$\underset{x→1}{\lim}\frac{x\sin\{x\}}{x-1}$, where {x} denotes the fractional part of x, is equal to
Answer & explanation
Correct answer: option 4
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