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NDA 1 2025 Maths, question 83

Topic: Differential calculus

Passage · Q83 to Q84

Let the function f(x)=sin⁡[x]f(x) = \sin\lbrack x\rbrack, where [⋅]\lbrack \cdot \rbrack is the greatest integer function and g(x)=∣x∣g(x) = |x|.

Q83

What is lim⁡x→0{f(x)g(x)}\lim_{x \rightarrow 0}{\{ f(x)g(x)\}} equal to?

  1. (a) -1
  2. (b) 0Correct answer
  3. (c) 1
  4. (d) Limit does not exist

Solution.

Answer: (b) 0

  1. Since ∣sin⁡[x]∣≤1\left| \sin\lbrack x\rbrack \right| \leq 1, we have ∣f(x)g(x)∣≤∣x∣\left| f(x)g(x) \right| \leq |x|.
  2. The right side tends to zero, so the squeeze theorem gives the limit 0.