JEE Main 2024 — Limits Question with Solution

Given: $a=\underset{x\to 0}{\lim }\frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$

$\Rightarrow a=\underset{x\to 0}{\lim }\frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}\times \frac{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}}$

$\Rightarrow a=\underset{x\to 0}{\lim }\frac{1+\sqrt{1+x^4}-2}{x^4\left(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}\right)}$

$\Rightarrow a=\underset{x\to 0}{\lim }\frac{\sqrt{1+x^4}-1}{x^4\left(\sqrt{1+\sqrt{1+0}}+\sqrt{2}\right)}$

$\Rightarrow a=\underset{x\to 0}{\lim }\frac{\sqrt{1+x^4}-1}{x^4\left(\sqrt{1+1}+\sqrt{2}\right)}$

$\Rightarrow a=\frac{1}{2\sqrt{2}}\underset{x\to 0}{\lim }\frac{\sqrt{1+x^4}-1}{x^4}$

$\Rightarrow a=\frac{1}{2\sqrt{2}}\underset{x\to 0}{\lim }\frac{\sqrt{1+x^4}-1}{x^4}\times \frac{\sqrt{1+x^4}+1}{\sqrt{1+x^4}+1}$

$\Rightarrow a=\frac{1}{2\sqrt{2}}\underset{x\to 0}{\lim }\frac{1+x^4-1}{x^4\left(\sqrt{1+x^4}+1\right)}$

$\Rightarrow a=\frac{1}{2\sqrt{2}}\underset{x\to 0}{\lim }\frac{x^4}{x^4\left(\sqrt{1+0}+1\right)}$

$\Rightarrow a=\frac{1}{2\sqrt{2}}\underset{x\to 0}{\lim }\frac{x^4}{x^4\left(2\right)}$

$\Rightarrow a=\frac{1}{4\sqrt{2}}\underset{x\to 0}{\lim }\frac{x^4}{x^4}$

$\Rightarrow a=\frac{1}{4\sqrt{2}}$

Also, $b=\underset{x\to 0}{\lim }\frac{{\sin }^{2}x}{\sqrt{2}-\sqrt{1+\cos x}}$

$\Rightarrow b=\underset{x\to 0}{\lim }\frac{\left(1-{\cos }^{2}x\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)}$

$\Rightarrow b=\underset{x\to 0}{\lim }\frac{\left(1-\cos x\right)\left(1+\cos x\right)(\sqrt{2}+\sqrt{1+\cos x})}{\left(1-\cos x\right)}$

$\Rightarrow b=\underset{x\to 0}{\lim }(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})$

$\Rightarrow b=(1+\cos 0)(\sqrt{2}+\sqrt{1+\cos 0})$

$\Rightarrow b=(1+1)(\sqrt{2}+\sqrt{1+1})$

$\Rightarrow b=\left(2\right)\left(2\sqrt{2}\right)$

$\Rightarrow b=4\sqrt{2}$

$\Rightarrow a{b}^3=\frac{1}{4\sqrt{2}}\times (4\sqrt{2}{)}^3$

$\Rightarrow a{b}^3=32$