If a = lim x→0 {√{1+√{1+x^4}} - √2} / x^4 and b = lim x→0 sin^2x / √2 - √{1+cosx}, then the value of ab^3 is:

Question

If \(a=\lim\limits _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^{4}}}-\sqrt{2}}{x^{4}}\) and \(b=\lim\limits _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{1+\cos x}}\), then the value of \(a b^{3}\) is :

(1) 36

(2) 32

(3) 25

(4) 30

Answer 1

Correct option is (2) 32

\(\mathrm{a}=\lim\limits _{\mathrm{x} \rightarrow 0} \frac{\sqrt{1+\sqrt{1+\mathrm{x}^{4}}}-\sqrt{2}}{\mathrm{x}^{4}}\)

\(=\lim \limits_{x \rightarrow 0} \frac{\sqrt{1+x^{4}}-1}{x^{4}\left(\sqrt{1+\sqrt{1+x^{4}}}+\sqrt{2}\right)}\)

\(=\lim \limits_{x \rightarrow 0} \frac{x^{4}}{x^{4}\left(\sqrt{1+\sqrt{1+x^{4}}}+\sqrt{2}\right)\left(\sqrt{1+x^{4}}+1\right)}\)

Applying limit \(\mathrm{a}=\frac{1}{4 \sqrt{2}}\)

\(b=\lim\limits _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{1+\cos x}}\)

\(=\lim\limits _{x \rightarrow 0} \frac{\left(1-\cos ^{2} x\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)}\)

\(b=\lim\limits _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})\)

Applying limits \(b=2(\sqrt{2}+\sqrt{2})=4 \sqrt{2}\)

Now, \(\mathrm{ab}^{3}=\frac{1}{4 \sqrt{2}} \times(4 \sqrt{2})^{3}=32\)