TS EAMCET · Maths · Limits
If \([t]\) represents the greatest integer \(\leq t\) then the value of \(\lim _{x \rightarrow 3} \frac{11-[2-x]}{[x+10]}\) is
- A 1
- B 8
- C 5
- D does not exist
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 3^-} \frac{11-[2-x]}{[x+10]} = \frac{11-[-1]}{[12]} = \frac{11-(-1)}{12} = \frac{12}{12} = 1\) \(\lim _{x \rightarrow 3^+} \frac{11-[2-x]}{[x+10]} = \frac{11-[-2]}{[13]} = \frac{11-(-2)}{13} = \frac{13}{13} = 1\) Limit = \(1\)
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