Assuming |x| to be so small that x^2 and high | vedclass

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(B) Using the binomial expansion $(1+x)^n \approx 1+nx$ for small $|x|$:$\sqrt{1+x} = (1+x)^{1/2} \approx 1+\frac{1}{2}x$$(1-x)^{3/2} \approx 1-\frac{

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$1-\frac{5x}{4}$

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(B) Using the binomial expansion $(1+x)^n \approx 1+nx$ for small $|x|$:$\sqrt{1+x} = (1+x)^{1/2} \approx 1+\frac{1}{2}x$$(1-x)^{3/2} \approx 1-\frac{3}{2}x$Substituting these into the expression:$\frac{(1+\frac{1}{2}x) + (1-\frac{3}{2}x)}{(1+x) + (1+\frac{1}{2}x)} = \frac{2-x}{2+\frac{3}{2}x} = \frac{2-x}{\frac{4+3x}{2}} = \frac{2(2-x)}{4+3x}$$= \frac{4-2x}{4+3x} = (4-2x)(4+3x)^{-1} = (4-2x) \cdot \frac{1}{4}(1+\frac{3}{4}x)^{-1}$$\approx \frac{1}{4}(4-2x)(1-\frac{3}{4}x) = \frac{1}{4}(4 - 3x - 2x + \frac{6}{4}x^2)$Neglecting $x^2$ terms:$\approx \frac{1}{4}(4-5x) = 1-\frac{5x}{4}$

Assuming $|x|$ to be so small that $x^2$ and higher powers of $x$ can be neglected,then $\frac{\sqrt{1+x}+(1-x)^{3/2}}{(1+x)+\sqrt{1+x}} = $

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