If $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\sin \left( {1 + \left[ x \right]} \right)}}{{\left[ x \right]}},{\rm{for}}\,\,\,\left[ x \right] \ne 0}\\ {0,\,\,\,\,\,\,\,\,\,\,{\rm{for}}\,\,\,\,\,\left[ x \right] = 0} \end{array}} \right.$ Where $\left[ x \right]$ denotes the greatest integer not exceeding $x$, then $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)$ is equal to

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Understanding +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sin \left( {1 + \left[ x \right]} \right)}}{{\left[ x \right]}},\mathrm{for}\,\,\,\left[ x \right] \ne 0}\\ {0,\,\,\,\,\,\,\,\,\,\,\mathrm{for}\,\,\,\,\,\left[ x \right] = 0} \end{array}} \right.

Where

\left[ x \right]

denotes the greatest integer not exceeding

x

, then

\lim\limits_{x \to {0^ - }} f\left( x \right)

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because as x approaches 0 from the left, [x] approaches -1, and the expression simplifies to sin(0)/-1 which equals 0.