If $f(x) = \left\{ \begin{array}{r} \frac{\sin{(1 + \lbrack x brack)}}{\lbrack x brack},\ \ \ \ for\ \lbrack x brack
eq 0 \ 0,\ \ \ \ \ for\ \lbrack x brack = 0 \end{array} ight.\$ Where $\lbrack x brack$ denotes the greatest integer not exceeding $x$, then $\lim_{x ightarrow 0^{-}}{f(x)}$ is equal to

Question

If $f(x) = \left\{ \begin{array}{r} \frac{\sin{(1 + \lbrack xbrack)}}{\lbrack xbrack},\ \ \ \ for\ \lbrack xbrack 
eq 0 \ 0,\ \ \ \ \ for\ \lbrack xbrack = 0 \end{array} ight.\$ Where $\lbrack xbrack$ denotes the greatest integer not exceeding $x$, then $\lim_{x ightarrow 0^{-}}{f(x)}$ is equal to

Dr. Abubacker · Accepted Answer

(b)
$\lim_{x \rightarrow 0^{-}}{f(x)} = \lim_{x \rightarrow 0^{-}}\frac{\sin{(1 + \lbrack x\rbrack)}}{\lbrack x\rbrack}$
$= \frac{\sin{(1 - 1)}}{- 1} = 0$

Explanation

Reviewed by Dr. Abubacker

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