JEE Main 2026 — Limits Question with Solution

Let $x-2=t$. As $x\to 2$, $t\to 0$.

The given limit can be written as:

${\lim }_{t\to 0}\frac{\sin ((t+2{)}^3-5(t+2{)}^2+a(t+2)+b)}{(\sqrt{t+1}-1){\log }_e(t+1)}=m$

The denominator can be approximated for small $t$ as:

${\lim }_{t\to 0}\frac{\sqrt{t+1}-1}{t}\cdot \frac{{\log }_e(t+1)}{t}\cdot t^2=\frac{1}{2}\cdot 1\cdot t^2=\frac{t^2}{2}$

For the limit to be finite, the argument of the sine function in the numerator must have $t^2$ as its lowest degree term in its expansion.

Let $P(t)=(t+2{)}^3-5(t+2{)}^2+a(t+2)+b$

$P(t)=t^3+6t^2+12t+8-5(t^2+4t+4)+at+2a+b$

$P(t)=t^3+t^2+(a-8)t+(2a+b-12)$

For the limit to exist, the constant and linear terms must be zero:

$a-8=0\Rightarrow a=8$

$2a+b-12=0\Rightarrow 16+b-12=0\Rightarrow b=-4$

Substituting $a$ and $b$ back into $P(t)$:

$P(t)=t^3+t^2$

The limit becomes:

$m={\lim }_{t\to 0}\frac{\sin (t^3+t^2)}{t^2/2}$

$m={\lim }_{t\to 0}\frac{\sin (t^2(t+1))}{t^2(t+1)}\cdot \frac{t^2(t+1)}{t^2/2}=1\cdot 2=2$

Therefore, $a+b+m=8-4+2=6$.

Answer: $6$