Limits Question 3
Question 3 - 2024 (27 Jan Shift 1)
If $a=\lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^{4}}}-\sqrt{2}}{x^{4}}$ and $b=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{1+\cos x}}$, then the value of $a b^{3}$ is :
(1) 36
(2) 32
(3) 25
(4) 30
Show Answer
Answer (2)
Solution
$$ \begin{aligned} a & =\lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^{4}}}-\sqrt{2}}{x^{4}} \\ & =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{4}}-1}{x^{4}\left(\sqrt{1+\sqrt{1+x^{4}}}+\sqrt{2}\right)} \end{aligned} $$
$$ =\lim _{x \rightarrow 0} \frac{x^{4}}{x^{4}\left(\sqrt{1+\sqrt{1+x^{4}}}+\sqrt{2}\right)\left(\sqrt{1+x^{4}}+1\right)} $$
Applying limit $a=\frac{1}{4 \sqrt{2}}$
$b=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{1+\cos x}}$
$=\lim _{x \rightarrow 0} \frac{\left(1-\cos ^{2} x\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)}$
$b=\lim _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})$
Applying limits $b=2(\sqrt{2}+\sqrt{2})=4 \sqrt{2}$
Now, $ab^{3}=\frac{1}{4 \sqrt{2}} \times(4 \sqrt{2})^{3}=32$
