JEE Main 2026 Jan 22 Shift 2, Mathematics Q5: Methods to Evaluate Limits

Expand numerator and denominator to order $x^2$:

Numerator $=\left[1+(a-1)x+\frac{(a-1{)}^2{x}^2}{2}\right]+2\left[1-\frac{b^2{x}^2}{2}\right]+(c-2)\left[1-x+\frac{x^2}{2}\right]$

$=(1+2+c-2)+x(a-1-c+2)+x^2\left[\frac{(a-1{)}^2}{2}-b^2+\frac{c-2}{2}\right]$.

Denominator $=x\left(1-\frac{x^2}{2}\right)-\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\dots \right)=\frac{x^2}{2}-\frac{x^3}{2!}+\dots$, leading term $\frac{x^2}{2}$.

For the limit to be finite, constant and linear terms in the numerator must vanish:

$c+1=0\Rightarrow c=-1$; $a-c+1=0\Rightarrow a=-2$.

Equating the $x^2$ coefficient: $\frac{(a-1{)}^2}{2}-b^2+\frac{c-2}{2}=1$, i.e., $\frac{9}{2}-b^2-\frac{3}{2}=1\Rightarrow b^2=2$.

$a^2+b^2+c^2=4+2+1=7$.