JEE Main 2026 — Limits Question with Solution

Using the standard expansions for small $x$, $\cos \theta \approx 1-\frac{{\theta }^2}{2}$ and $\sin \theta \approx \theta$.

The given limit can be written as:

$\underset{x\to 0}{\lim }\frac{1-\left(1-\frac{{\alpha }^2{x}^2}{2}\right)\left(1-\frac{(\alpha +1{)}^2{x}^2}{2}\right)\left(1-\frac{(\alpha +2{)}^2{x}^2}{2}\right)}{((\alpha +1)x{)}^2}$

Neglecting higher powers of $x$, the numerator simplifies to:

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

Substituting this back into the limit:

$\underset{x\to 0}{\lim }\frac{\frac{x^2}{2}\left({\alpha }^2+(\alpha +1{)}^2+(\alpha +2{)}^2\right)}{(\alpha +1{)}^2{x}^2}=\frac{{\alpha }^2+(\alpha +1{)}^2+(\alpha +2{)}^2}{2(\alpha +1{)}^2}$

We are given that this limit is equal to $2$. Therefore:

$\frac{{\alpha }^2+(\alpha +1{)}^2+(\alpha +2{)}^2}{2(\alpha +1{)}^2}=2$

${\alpha }^2+(\alpha +1{)}^2+(\alpha +2{)}^2=4(\alpha +1{)}^2$

Let $y=\alpha +1$. Then $\alpha =y-1$ and $\alpha +2=y+1$. The equation becomes:

$(y-1{)}^2+y^2+(y+1{)}^2=4y^2$

$y^2-2y+1+y^2+y^2+2y+1=4y^2$

$3y^2+2=4y^2$

$y^2=2$

Substituting $y=\alpha +1$ back into the equation:

$(\alpha +1{)}^2=2$

${\alpha }^2+2\alpha +1=2$

${\alpha }^2+2\alpha -1=0$

The roots of this quadratic equation represent all possible values of $\alpha$. The product of the roots is given by $\frac{c}{a}$:

$\text{Product}=\frac{-1}{1}=-1$

Answer: $-1$