If |x| is so small that x^2 and higher powers | vedclass

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(B) Using binomial expansion $(1+u)^n \approx 1+nu$ for small $u$, we have: $\sqrt{4+x} = 2(1+\frac{x}{4})^{\frac{1}{2}} \approx 2(1+\frac{x}{8}) = 2+

Vedclass · Accepted Answer

$5$

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(B) Using binomial expansion $(1+u)^n \approx 1+nu$ for small $u$, we have: $\sqrt{4+x} = 2(1+\frac{x}{4})^{\frac{1}{2}} \approx 2(1+\frac{x}{8}) = 2+\frac{x}{4}$ $\sqrt[3]{8-x} = 2(1-\frac{x}{8})^{\frac{1}{3}} \approx 2(1-\frac{x}{24}) = 2-\frac{x}{12}$ $(1-\frac{2x}{3})^{-\frac{3}{2}} \approx 1+(-\frac{3}{2})(-\frac{2x}{3}) = 1+x$ Substituting these into the expression: $\frac{(2+\frac{x}{4})+(2-\frac{x}{12})}{1} 	imes (1+x) = (4+\frac{3x-x}{12})(1+x) = (4+\frac{x}{6})(1+x)$ Neglecting $x^2$ terms: $4+4x+\frac{x}{6} = 4+\frac{25x}{6}$ Substituting $x=\frac{6}{25}$: $4+\frac{25}{6} 	imes \frac{6}{25} = 4+1 = 5$

If $|x|$ is so small that $x^2$ and higher powers of $x$ may be neglected, then the approximate value of $\frac{\sqrt{4+x}+\sqrt[3]{8-x}}{\left(1-\frac{2x}{3}\right)^{\frac{3}{2}}}$ when $x=\frac{6}{25}$ is

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