Q221 / 270 · NDA 2025 (Phase 2) · Mathematics
For the following two (02) items:
Let $f(x) = \begin{cases} \dfrac{1 - \cos 2x}{x^2}, & x < 0 \\ 9, & x = 0 \\ \dfrac{\sqrt{x}}{\sqrt{(16 + \sqrt{x})} - 4}, & x > 0 \end{cases}$
What is $\lim_{x \to 0^-} f(x)$ equal to?
Let $f(x) = \begin{cases} \dfrac{1 - \cos 2x}{x^2}, & x < 0 \\ 9, & x = 0 \\ \dfrac{\sqrt{x}}{\sqrt{(16 + \sqrt{x})} - 4}, & x > 0 \end{cases}$
What is $\lim_{x \to 0^-} f(x)$ equal to?
- Correct answerA$2$
- B$4$
- C$6$
- D$8$
Solution
For $x < 0$, the function is given by $f(x) = \dfrac{1 - \cos 2x}{x^2}$
To find the left-hand limit, we evaluate $\lim_{x \to 0^-} f(x)$
$\lim_{x \to 0^-} \dfrac{1 - \cos 2x}{x^2}$
Using the trigonometric identity $1 - \cos 2x = 2 \sin^2 x$
$\lim_{x \to 0^-} \dfrac{2 \sin^2 x}{x^2} = 2 \left( \lim_{x \to 0^-} \dfrac{\sin x}{x} \right)^2$
Since $\lim_{x \to 0} \dfrac{\sin x}{x} = 1$, we get:
$2 \times (1)^2 = 2$
Answer: $2$
To find the left-hand limit, we evaluate $\lim_{x \to 0^-} f(x)$
$\lim_{x \to 0^-} \dfrac{1 - \cos 2x}{x^2}$
Using the trigonometric identity $1 - \cos 2x = 2 \sin^2 x$
$\lim_{x \to 0^-} \dfrac{2 \sin^2 x}{x^2} = 2 \left( \lim_{x \to 0^-} \dfrac{\sin x}{x} \right)^2$
Since $\lim_{x \to 0} \dfrac{\sin x}{x} = 1$, we get:
$2 \times (1)^2 = 2$
Answer: $2$
Hence, the correct answer is option: 'A'.