mathop {lim }limits_{y to 0} frac{{sqrt {1 + | vedclass

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Question

(A) Let $u = y^4$. As $y 	o 0$,$u 	o 0$. The expression becomes $\mathop {\lim }\limits_{u 	o 0} \frac{{\sqrt {1 + \sqrt {1 + u} }  - \sqrt 2 }}{u}

Vedclass · Accepted Answer

exists and equals $\frac{1}{{4\sqrt 2 }}$

Vedclass · Answer

(A) Let $u = y^4$. As $y 	o 0$,$u 	o 0$. The expression becomes $\mathop {\lim }\limits_{u 	o 0} \frac{{\sqrt {1 + \sqrt {1 + u} }  - \sqrt 2 }}{u}$.Using the binomial expansion $(1+x)^n \approx 1+nx$ for small $x$,we have $\sqrt{1+u} \approx 1 + \frac{u}{2}$.Substituting this into the expression: $\sqrt{1 + (1 + \frac{u}{2})} = \sqrt{2 + \frac{u}{2}} = \sqrt{2} \sqrt{1 + \frac{u}{4}}$.Using the expansion again: $\sqrt{2} (1 + \frac{u}{8}) = \sqrt{2} + \frac{\sqrt{2}u}{8}$.Now,the limit is $\mathop {\lim }\limits_{u 	o 0} \frac{{\sqrt{2} + \frac{\sqrt{2}u}{8} - \sqrt{2}}}{u} = \mathop {\lim }\limits_{u 	o 0} \frac{\sqrt{2}}{8} = \frac{\sqrt{2}}{8} = \frac{1}{4\sqrt{2}}$.

$\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 }}{{{y^4}}} = $

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