Let $f\left(x\right)=x{(-1)}^{\left[\frac{1}{x}\right]}.x\ne 0$, where [x] denotes the greatest integer less than or equal to x. then $\underset{x\to 0}{lim}f\left(x\right)$

The correct option is C

is equal to 0

Since [1/x] is an integer .
${(-1)}^{\left[\frac{1}{x}\right]}=\pm 1$

Hence f(x)=$\pm x$
And $\underset{x\to 0}{lim}f\left(x\right)$ = 0