lim _{x rightarrow 0} frac{sqrt{1+sqrt{1+x^4} | vedclass

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Question

(A) To evaluate the limit $L = \lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2+x^5+x^6}}{x^4}$,we use rationalization.Multiply the numerat

Vedclass · Accepted Answer

$\frac{1}{4 \sqrt{2}}$

Vedclass · Answer

(A) To evaluate the limit $L = \lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2+x^5+x^6}}{x^4}$,we use rationalization.Multiply the numerator and denominator by the conjugate $\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6}$:$L = \lim _{x \rightarrow 0} \frac{(1+\sqrt{1+x^4})-(2+x^5+x^6)}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6})}$$L = \lim _{x \rightarrow 0} \frac{\sqrt{1+x^4}-1-x^5-x^6}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2+x^5+x^6})}$Using the binomial expansion $\sqrt{1+u} \approx 1 + \frac{u}{2}$ for small $u$,where $u = x^4$,we have $\sqrt{1+x^4} \approx 1 + \frac{x^4}{2}$.Substituting this into the limit:$L = \lim _{x \rightarrow 0} \frac{1 + \frac{x^4}{2} - 1 - x^5 - x^6}{x^4(\sqrt{1+1} + \sqrt{2})} = \lim _{x \rightarrow 0} \frac{\frac{x^4}{2} - x^5 - x^6}{x^4(2\sqrt{2})}$$L = \lim _{x \rightarrow 0} \frac{\frac{1}{2} - x - x^2}{2\sqrt{2}} = \frac{1/2}{2\sqrt{2}} = \frac{1}{4\sqrt{2}}$

$\lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2+x^5+x^6}}{x^4} = $

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