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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Limits

NDA 2025 · Mathematics · Limits

For the following two (02) items: Let f(x) = cases (1 - 2x)/(x^2),...

For the following two (02) items:
Let f(x)=f(x) =\begin{cases} \dfrac{1 - \cos 2x}{x^2}, & x < 0 \\ 9, & x = 0 \\ \dfrac{\sqrt{x}}{\sqrt{(16 + \sqrt{x})} - 4}, & x > 0 \end{cases}$<br><br>Whatis<br><br>What is\lim_{x \to 0^+} f(x)$ equal to?

Options

  1. A66
  2. B77
  3. C88
  4. D99

Correct answer

C. 88

Step-by-step solution

For x>0x > 0, the function is given by f(x)=x16+x−4f(x) = \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}

To find the right-hand limit, we evaluate lim⁡x→0+f(x)\lim_{x \to 0^+} f(x)

lim⁡x→0+x16+x−4\lim_{x \to 0^+} \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}

Rationalizing the denominator by multiplying the numerator and the denominator by 16+x+\sqrt{16 + \sqrt{x}} +4:<br><br>lim⁡x→0+x(16+x+4)(16+x−4)(16+x+4):<br><br>\lim_{x \to 0^+} \dfrac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{(\sqrt{16 + \sqrt{x}} - 4)(\sqrt{16 + \sqrt{x}} + 4)}

lim⁡x→0+x(16+x+4)16+x−16\lim_{x \to 0^+} \dfrac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{16 + \sqrt{x} - 16}

lim⁡x→0+x(16+x+4)x\lim_{x \to 0^+} \dfrac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{\sqrt{x}}

Canceling xfromthenumeratorandthedenominator:<br><br>lim⁡x→0+(16+x+4)\sqrt{x} from the numerator and the denominator:<br><br>\lim_{x \to 0^+} (\sqrt{16 + \sqrt{x}} + 4)

Substituting x=0x = 0:

16+0+\sqrt{16 + 0} +4 = 4 + 4 = 8<br><br>Answer:8<br><br>Answer: 8
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