If $\mathrm{x}$ is so small that $\mathrm{x}^3$ and higher powers of $\mathrm{x}$ may be neglected, then $\frac{(1+x)^{3 / 2}-\left(1+\frac{1}{2} x\right)^3}{(1-x)^{1 / 2}}$

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Aicharya · Accepted Answer

$ \begin{aligned} & (1-x)^{1 / 2}\left[1+\frac{3}{2} x+\frac{3}{2}\left(\frac{3}{2}-1ight) x^2-1-3\left(\frac{1}{2} xight)-3(2)\left(\frac{1}{2} xight)^2ight] \ & =(1-x)^{1 / 2}\left[-\frac{3}{8} x^2ight]=-\frac{3}{8} x^2 \end{aligned} $

If $\mathrm{x}$ is so small that $\mathrm{x}^3$ and higher powers of $\mathrm{x}$ may be neglected, then…

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