TS EAMCET · Maths · Limits
If \([x]\) is the greatest integer function then \(\lim _{x \rightarrow 3^{-}} \frac{(3-|x|+\sin |3-x|) \cos [9-3 x]}{|3-x|[3 x-9]}=\)
- A \(0\)
- B \(1\)
- C \(2\)
- D \(-2\)
Answer & Solution
Correct Answer
(D) \(-2\)
Step-by-step Solution
Detailed explanation
As \(x \rightarrow 3^{-}\): \(|x|=x\), \(|3-x|=3-x\), \([9-3x]=0\), \([3x-9]=-1\). \(\lim _{x \rightarrow 3^{-}} \frac{(3-x+\sin (3-x)) \cdot 1}{(3-x)(-1)}\) \(= \lim _{x \rightarrow 3^{-}} -\left( \frac{3-x}{3-x} + \frac{\sin (3-x)}{3-x} \right)\) \(= -(1 + 1)\) \(= -2\)
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