JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Limits

Expand the numerator using Taylor series around $x=0$: $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)$, $e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+O(x^4)$, $\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)$. The numerator becomes $a(1+x+\frac{x^2}{2}+\frac{x^3}{6})-b(1-\frac{x^2}{2})+c(1-x+\frac{x^2}{2}-\frac{x^3}{6})=(a-b+c)+(a-c)x+\frac{1}{2}(a+b+c)x^2+\frac{1}{6}(a-c)x^3+O(x^4)$. The denominator is $x\sin x=x(x-\frac{x^3}{6}+O(x^5))=x^2-\frac{x^4}{6}+O(x^6)$. For the limit to exist and equal 2, we need: (1) $a-b+c=0$ (constant term vanishes), (2) $a-c=0$ (linear term vanishes), (3) $\frac{a+b+c}{2}=2$ (quadratic coefficient gives the limit). From (2): $a=c$. From (1): $a-b+a=0$, so $b=2a$. From (3): $a+2a+a=4$, so $4a=4$, giving $a=1$, $b=2$, $c=1$. Therefore $a+b+c=4$.