If f(x)=sin(1+[x])[x] for [x]≠00 for [x]=0where[x] denotes the greatest integer ≤x then limx→0-f(x) equals

For -1<x<0,[x]=-1, so limx→0-sin(1+[x])[x]=sin0-1=0

If f(x)=sin(1+[x])[x] for [x]≠00 for [x]=0where[x] denotes the greatest integer ≤x then limx→0-f(x) equals

For -1<x<0,[x]=-1, so limx→0-sin(1+[x])[x]=sin0-1=0

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If $f (x) = \{\begin{array}{cl} \frac{\sin (1 + [x])}{[x]} & for [x] \neq 0 \\ 0 & for [x] = 0 \end{array}$ where
[x] denotes the greatest integer $\leq x$ then $\lim_{x \to 0 -} f (x)$ equals

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For $- 1 < x < 0, [x] = - 1, so$

$\lim_{x \to 0 -} \frac{\sin (1 + [x])}{[x]} = \frac{\sin 0}{- 1} = 0$

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If f(x)=sin(1+[x])[x] for [x]≠00 for [x]=0where[x] denotes the greatest integer ≤x then limx→0-f(x) equals

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If $f (x) = \{\begin{array}{cl} \frac{\sin (1 + [x])}{[x]} & for [x] \neq 0 \\ 0 & for [x] = 0 \end{array}$ where
[x] denotes the greatest integer $\leq x$ then $\lim_{x \to 0 -} f (x)$ equals