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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Limits

NDA 2025 · Mathematics · Limits

Consider the following for the two (02) items that follow: Let the...

Consider the following for the two (02) items that follow:
Let the function f(x)=x2+9f(x) = x^2 + 9.

What is lim⁡x→0f(x)−3f(x)+7−4\lim_{x \to 0} \dfrac{\sqrt{f(x)} - 3}{\sqrt{f(x)+7} - 4} equal to?

Options

  1. A23\dfrac{2}{3}
  2. B11
  3. C43\dfrac{4}{3}
  4. D22

Correct answer

C. 43\dfrac{4}{3}

Step-by-step solution

Given f(x)=x2+9f(x) = x^2 + 9.

The limit is lim⁡x→0x2+9−3x2+16−4\lim_{x \to 0} \dfrac{\sqrt{x^2 + 9} - 3}{\sqrt{x^2 + 16} - 4}.

Rationalizing the numerator and the denominator:

lim⁡x→0(x2+9−3)(x2+9+3)(x2+16−4)(x2+16+4)×x2+16+4x2+9+3\lim_{x \to 0} \dfrac{(\sqrt{x^2 + 9} - 3)(\sqrt{x^2 + 9} + 3)}{(\sqrt{x^2 + 16} - 4)(\sqrt{x^2 + 16} + 4)} \times \dfrac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}

lim⁡x→0x2+9−9x2+16−16×x2+16+4x2+9+3\lim_{x \to 0} \dfrac{x^2 + 9 - 9}{x^2 + 16 - 16} \times \dfrac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}

lim⁡x→0x2x2×x2+16+4x2+9+3\lim_{x \to 0} \dfrac{x^2}{x^2} \times \dfrac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}

Cancelling x2x^2 and substituting x=0x = 0:

16+49+3=4+43+3=86=43\dfrac{\sqrt{16} + 4}{\sqrt{9} + 3} = \dfrac{4 + 4}{3 + 3} = \dfrac{8}{6} = \dfrac{4}{3}

Answer: 43\dfrac{4}{3}
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