NDA 1 2025 Maths, question 93

Topic: Differential calculus

Passage · Q93 to Q94

Let the function $f(x) = x^{2} + 9$ .

Q93

What is $\lim_{x \rightarrow 0}\frac{\sqrt{f(x)} - 3}{\sqrt{f(x) + 7} - 4}$ equal to?

Solution.

Answer: (c) 4/3

Rationalise both differences.
For $x \neq 0$ , the quotient becomes $\frac{x^{2}/\left( \sqrt{x^{2} + 9} + 3 \right)}{x^{2}/\left( \sqrt{x^{2} + 16} + 4 \right)} = \frac{\sqrt{x^{2} + 16} + 4}{\sqrt{x^{2} + 9} + 3}$ .
Taking $x \rightarrow 0$ now gives $8/6 = 4/3$ .