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MHT CET · Maths · Limits

For each \(x \in \mathbb{R}\), Let \([x]\) represent greatest integer function, then \(\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}\) is equal to

  1. A 0
  2. B 1
  3. C \(\sin 1\)
  4. D \(-\sin 1\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-\sin 1\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{x} \\ & \text { For } x \rightarrow 0^{-},[x]=-1,|x|=-x \\ \therefore \quad & \lim _{x \rightarrow 0^{-}} \frac{x(-1-x) \sin (-1)}{-x} \\ & =\lim _{x \rightarrow 0^{-}} \frac{-(1+x) \sin (1)}{1} \quad \ldots[x \rightarrow 0, x \neq 0] \\ \therefore & =-(1+0) \sin 1 \\ & =-\sin 1\end{aligned}\)