Q242 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
Let the function $f(x) = x^2 + 9$.
What is $\lim_{x \to 0} \dfrac{\sqrt{f(x)} - 3}{\sqrt{f(x)+7} - 4}$ equal to?
Let the function $f(x) = x^2 + 9$.
What is $\lim_{x \to 0} \dfrac{\sqrt{f(x)} - 3}{\sqrt{f(x)+7} - 4}$ equal to?
- A$\dfrac{2}{3}$
- B$1$
- Correct answerC$\dfrac{4}{3}$
- D$2$
Solution
Given $f(x) = x^2 + 9$.
The limit is $\lim_{x \to 0} \dfrac{\sqrt{x^2 + 9} - 3}{\sqrt{x^2 + 16} - 4}$.
Rationalizing the numerator and the denominator:
$\lim_{x \to 0} \dfrac{(\sqrt{x^2 + 9} - 3)(\sqrt{x^2 + 9} + 3)}{(\sqrt{x^2 + 16} - 4)(\sqrt{x^2 + 16} + 4)} \times \dfrac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}$
$\lim_{x \to 0} \dfrac{x^2 + 9 - 9}{x^2 + 16 - 16} \times \dfrac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}$
$\lim_{x \to 0} \dfrac{x^2}{x^2} \times \dfrac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}$
Cancelling $x^2$ and substituting $x = 0$:
$\dfrac{\sqrt{16} + 4}{\sqrt{9} + 3} = \dfrac{4 + 4}{3 + 3} = \dfrac{8}{6} = \dfrac{4}{3}$
Answer: $\dfrac{4}{3}$
The limit is $\lim_{x \to 0} \dfrac{\sqrt{x^2 + 9} - 3}{\sqrt{x^2 + 16} - 4}$.
Rationalizing the numerator and the denominator:
$\lim_{x \to 0} \dfrac{(\sqrt{x^2 + 9} - 3)(\sqrt{x^2 + 9} + 3)}{(\sqrt{x^2 + 16} - 4)(\sqrt{x^2 + 16} + 4)} \times \dfrac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}$
$\lim_{x \to 0} \dfrac{x^2 + 9 - 9}{x^2 + 16 - 16} \times \dfrac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}$
$\lim_{x \to 0} \dfrac{x^2}{x^2} \times \dfrac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3}$
Cancelling $x^2$ and substituting $x = 0$:
$\dfrac{\sqrt{16} + 4}{\sqrt{9} + 3} = \dfrac{4 + 4}{3 + 3} = \dfrac{8}{6} = \dfrac{4}{3}$
Answer: $\dfrac{4}{3}$
Hence, the correct answer is option: 'C'.