NIMCET 2026 — Mathematics PYQ

To solve this limit, we can use the Maclaurin series expansion for $\cos x$ and $\sin x$ near $x=0$.

1. Expansion formulas:

• $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots =1-\frac{x^2}{2}+\frac{x^4}{24}-\dots$

• $\sin x=x-\frac{x^3}{3!}+\cdots =x-\frac{x^3}{6}+\dots$

2. Substitute the series into the numerator:

Numerator $=x^2+2\cos x-2$

$$=x^2+2\left(1-\frac{x^2}{2}+\frac{x^4}{24}-\dots \right)-2$$

$$=x^2+2-x^2+\frac{2x^4}{24}-\dots$$

$$=\frac{x^4}{12}+\dots$$

3. Simplify the denominator:

Denominator $=x{\sin }^{3}x$

Since $\sin x\approx x$ for small $x$:

$$x{\sin }^{3}x\approx x(x{)}^3=x^4$$

4. Evaluate the limit:

Substituting the leading terms back into the limit expression:

$$\underset{x\to 0}{\lim }\left(\frac{\frac{x^4}{12}}{x^4}\right)$$

$$\underset{x\to 0}{\lim }\left(\frac{1}{12}\right)=\frac{1}{12}$$

Correct Option: 1. $\frac{1}{12}$