NDA 2 2025 Maths, question 71

Topic: Differential calculus

Passage · Q71 to Q72

Let $f(x)=\left\{ \begin{matrix} \frac{1-\cos 2x}{x^{2}} & , x<0 \\ 9 & , x=0 \\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4} & , x>0 \end{matrix} \right.$

Q71

What is $\underset{x\to 0-}{\lim }f(x)$ equal to?

(a) $2$ Correct answer
(b) $4$
(c) $6$
(d) $8$

Solution.

Answer: (a) $2$

Use 1−cos2x=2sin²x.
Then (1−cos2x)/x²=2(sin x/x)²→2 as x approaches zero from the left.
The separately assigned value f(0)=9 does not determine this limit.