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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. A22
  2. B44
  3. C66
  4. D88

Correct answer

A. 22

Step-by-step solution

For x<0x < 0, the function is given by f(x)=1−cos⁡2xx2f(x) = \dfrac{1 - \cos 2x}{x^2}

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

lim⁡x→0−1−cos⁡2xx2\lim_{x \to 0^-} \dfrac{1 - \cos 2x}{x^2}

Using the trigonometric identity 1−cos⁡2x=2sin⁡2x1 - \cos 2x = 2 \sin^2 x

lim⁡x→0−2sin⁡2xx2=2(lim⁡x→0−sin⁡xx)2\lim_{x \to 0^-} \dfrac{2 \sin^2 x}{x^2} = 2 \left( \lim_{x \to 0^-} \dfrac{\sin x}{x} \right)^2

Since lim⁡x→0sin⁡xx=1\lim_{x \to 0} \dfrac{\sin x}{x} = 1, we get:

2×(1)2=22 \times (1)^2 = 2

Answer: 22
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