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UPSC NDA I 2025 Mathematics Paper

Question Numbers: 93-94

Consider the following for the two (02) items that follow:

Let the function

f(x) = x ^{2} + 9

Question : 93 of 120

Marks: +1, -0

What is

\underset{x→0}{lim}\frac{\sqrt{f(x)}-3}{\sqrt{f(x)+7}-4}

equal to?

Solution:

Concept:

Use conjugate multiplication to simplify limits involving square roots and cancel common factors.

Explanation:

Given

f(x) = \sqrt{x^{2}+9} - 3

and

g(x) = \sqrt{x^{2}+16} - 4

. We need

\underset{x\to 0}{\lim} \frac{f(x)}{g(x)}

Multiply numerator and denominator by their conjugates:

\frac{\sqrt{x^{2}+9}-3}{\sqrt{x^{2}+16}-4} \times \frac{\sqrt{x^{2}+9}+3}{\sqrt{x^{2}+9}+3} \times \frac{\sqrt{x^{2}+16}+4}{\sqrt{x^{2}+16}+4}

Numerator simplifies:

(\sqrt{x^{2}+9}-3)(\sqrt{x^{2}+9}+3) = x^{2}

Denominator simplifies:

(\sqrt{x^{2}+16}-4)(\sqrt{x^{2}+16}+4) = x^{2}

The expression becomes

\frac{x^{2}}{x^{2}} \times \frac{\sqrt{x^{2}+16}+4}{\sqrt{x^{2}+9}+3} = \frac{\sqrt{x^{2}+16}+4}{\sqrt{x^{2}+9}+3}

Take limit as

x\to 0

\frac{\sqrt{16}+4}{\sqrt{9}+3} = \frac{4+4}{3+3} = \frac{8}{6} = \frac{4}{3}

Answer:

\frac{4}{3}

(Option C).

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