A) \[-1\]
B) 1
C) 0
D) none of these
Correct Answer: C
Solution :
Key Idea: If \[x\]is any real number, then \[[x]\le x,\] Now, \[\underset{x\to 0}{\mathop{\lim }}\,[\cos x]\] When \[x\to {{0}^{-}},-\frac{\pi }{2}<x<0.\] \[0<\cos x<1\Rightarrow [\cos x]=0\] \[\therefore \] \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,[\cos x]=0\] When \[x\to {{0}^{+}},0<x<\frac{\pi }{2}\] \[\Rightarrow \] \[0<\cos x<1\]\[\Rightarrow \]\[[\cos x]=0\] \[\therefore \] \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,[\cos x]=0\] Hence, \[\underset{x\to 0}{\mathop{\lim }}\,[\cos x]=0\] Note: If any function LHL is equal to RHL, then the limit will exist otherwise not.
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