JEE Main 2025 Apr 4 Shift 1, Mathematics Q7: Methods to Evaluate Limits

Substitute $h=x-1\to 0^{+}$. The limit becomes $\underset{h\to 0}{\lim }\frac{h(6+\lambda \cos h)-\mu \sin h}{h^3}=-1$.

Use Taylor series: $\cos h=1-\frac{h^2}{2}+O(h^4)$ and $\sin h=h-\frac{h^3}{6}+O(h^5)$.

Numerator $=h\left(6+\lambda -\frac{\lambda h^2}{2}\right)-\mu \left(h-\frac{h^3}{6}\right)=(6+\lambda -\mu )h+\left(-\frac{\lambda }{2}+\frac{\mu }{6}\right)h^3+\cdots$.

For the limit to be finite, the coefficient of $h$ must vanish: $6+\lambda -\mu =0$, i.e. $\mu -\lambda =6$.

The coefficient of $h^3$ then equals $-1$: $-\frac{\lambda }{2}+\frac{\mu }{6}=-1$, i.e. $-3\lambda +\mu =-6$, so $\mu =3\lambda -6$.

From the two relations: $3\lambda -6-\lambda =6$, hence $\lambda =6$ and $\mu =12$. Therefore $\lambda +\mu =18$.