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

Question

If $a=\underset{x	o 0}{lim}\frac{\sqrt{1+\sqrt{1+{x}^{4}}}-\sqrt{2}}{{x}^{4}}$ and $b=\underset{x	o 0}{lim}\frac{{\sin }^{2}x}{\sqrt{2}-\sqrt{1+\cos x}}$, then the value of $a{b}^{3}$ is :

Accepted Answer

$\begin{gathered} a= \underset{x	o 0}{lim}\frac{\sqrt{1+\sqrt{1+{x}^{4}}}-\sqrt{2}}{{x}^{4}} \ =\underset{x	o 0}{lim}\frac{\sqrt{1+{x}^{4}}-1}{{x}^{4}(\sqrt{1+\sqrt{1+{x}^{4}}}+\sqrt{2})} \ =\underset{x	o 0}{lim}\frac{{x}^{4}}{{x}^{4}(\sqrt{1+\sqrt{1+{x}^{4}}}+\sqrt{2})(\sqrt{1+{x}^{4}}+1)} \end{gathered}$

Applying limit $a=\frac{1}{4\sqrt{2}}$

$\begin{gathered} b=\underset{x	o 0}{lim}\frac{{\sin }^{2}x}{\sqrt{2}-\sqrt{1+\cos x}} \ =\underset{x	o 0}{lim}\frac{(1-{\cos }^{2}x)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} \ b=\underset{x	o 0}{lim}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x}) \end{gathered}$

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

Now, $a{b}^{3}=\frac{1}{4\sqrt{2}}	imes (4\sqrt{2}{)}^{3}=32$

Solutions · 1