JEE Main 2024 (Online) 27th January Morning ShiftQuestion 1
If $a=\lim_{x\to0}\frac{\sqrt{1+\sqrt{1+x^{4}}}-\sqrt{2}}{x^{4}}$ and $b=\lim_{x\to0}\frac{\sin^{2}x}{\sqrt{2}-\sqrt{1+\cos x}}$, then the value of $ab^{3}$ is :
Limits & Continuity
Solution
$\sqrt{1+x^4}\approx1+\tfrac{x^4}{2}$, so $\sqrt{2+x^4/2}-\sqrt2\approx\frac{x^4/2}{2\sqrt2}$, giving $a=\frac{1}{4\sqrt2}$. $\sqrt{1+\cos x}=\sqrt{2-x^2/2+\dots}\approx\sqrt2-\frac{x^2}{4\sqrt2}$, so $\sqrt2-\sqrt{1+\cos x}\approx\frac{x^2}{4\sqrt2}$ and $b=4\sqrt2$. $ab^3=\frac{1}{4\sqrt2}\cdot128\sqrt2=32$. Option (D).