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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Limits

NTA Abhyas 2020 · Mathematics · Limits

The value of _ x → 0 (sin x)/(3)[(5)/(x)]is equal to (where, [.]...

The value of lim⁡x→0sinx3[5x]isequalto</p><p>(where,[.]represents\lim_{x \to 0}\dfrac{sin x}{3}\left[\dfrac{5}{x}\right]is equal to</p> <p>(where, \left[.\right] represents the greatest integer function)

Options

  1. A13\dfrac{1}{3}
  2. B00
  3. C53\dfrac{5}{3}
  4. D11

Correct answer

C. 53\dfrac{5}{3}

Step-by-step solution

5x−1<[5x]≤5x\dfrac{5}{x} - 1 < \left[\dfrac{5}{x}\right] \le \dfrac{5}{x}
sinx3(5x−1)⏟h(x)<sinx3[5x]⏟f(x)≤sinx3(5x)⏟g(x)\mathop{\mathop{\dfrac{sin x}{3}\left(\dfrac{5}{x} - 1\right)}_{⏟}}_{h\left(x\right)} < \mathop{\mathop{\dfrac{sin x}{3}\left[\dfrac{5}{x}\right]}_{⏟}}_{f\left(x\right)} \le \mathop{\mathop{\dfrac{sin x}{3}\left(\dfrac{5}{x}\right)}_{⏟}}_{g\left(x\right)}
by sandwich theorem
∵∵ lim⁡x→0g(x)=lim⁡x→0h(x)=53\lim_{x \to 0}g\left(x\right) = \lim_{x \to 0}h\left(x\right) = \dfrac{5}{3}
∴lim⁡x→0f(x)=53∴ \lim_{x \to 0}f\left(x\right) = \dfrac{5}{3}
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