UPSC NDA I 2025 Mathematics Paper

Given,
The function is f(x)=√‌x2+9−3 and g(x)=√‌x2+16−4.
We are tasked with finding:

lim

x⟶0

‌

f(x)

g(x)

Multiply both the numerator and denominator by their respective conjugates:
‌

√x2+9−3

√x2+16−4

×‌

√x2+9+3

√x2+9+3

×‌

√x2+16+4

√x2+16+4

Simplify the numerator:
(√‌x2+9−3)(√‌x2+9+3)=x2
Simplify the denominator:
(√‌x2+16−4)(√‌x2+16+4)=x2
Now, the expression becomes:
‌

x2

x2

×‌

√x2+16+4

√x2+9+3

Simplify and evaluate the limit:
‌

√x2+16+4

√x2+9+3

becomes:
‌

√16+4

√9+3

=‌

4+4

3+3

=‌

8

6

=‌

4

3