Q222 / 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?
- A$6$
- B$7$
- Correct answerC$8$
- D$9$
Solution
For $x > 0$, the function is given by $f(x) = \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}$
To find the right-hand limit, we evaluate $\lim_{x \to 0^+} f(x)$
$\lim_{x \to 0^+} \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}$
Rationalizing the denominator by multiplying the numerator and the denominator by $\sqrt{16 + \sqrt{x}} + 4$:
$\lim_{x \to 0^+} \dfrac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{(\sqrt{16 + \sqrt{x}} - 4)(\sqrt{16 + \sqrt{x}} + 4)}$
$\lim_{x \to 0^+} \dfrac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{16 + \sqrt{x} - 16}$
$\lim_{x \to 0^+} \dfrac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{\sqrt{x}}$
Canceling $\sqrt{x}$ from the numerator and the denominator:
$\lim_{x \to 0^+} (\sqrt{16 + \sqrt{x}} + 4)$
Substituting $x = 0$:
$\sqrt{16 + 0} + 4 = 4 + 4 = 8$
Answer: $8$
To find the right-hand limit, we evaluate $\lim_{x \to 0^+} f(x)$
$\lim_{x \to 0^+} \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}$
Rationalizing the denominator by multiplying the numerator and the denominator by $\sqrt{16 + \sqrt{x}} + 4$:
$\lim_{x \to 0^+} \dfrac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{(\sqrt{16 + \sqrt{x}} - 4)(\sqrt{16 + \sqrt{x}} + 4)}$
$\lim_{x \to 0^+} \dfrac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{16 + \sqrt{x} - 16}$
$\lim_{x \to 0^+} \dfrac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{\sqrt{x}}$
Canceling $\sqrt{x}$ from the numerator and the denominator:
$\lim_{x \to 0^+} (\sqrt{16 + \sqrt{x}} + 4)$
Substituting $x = 0$:
$\sqrt{16 + 0} + 4 = 4 + 4 = 8$
Answer: $8$
Hence, the correct answer is option: 'C'.