If $x$ is numerically so small so that $x^2$ and higher powers of $x$ can be neglected, then $\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32+5 x)^{-1 / 5}$ is approximately equal to

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Aicharya · Accepted Answer

$\begin{aligned} (1+ & \left.\frac{2 x}{3}ight)^{3 / 2}(32+5 x)^{-1 / 5} \ & =\left[1+\frac{3}{2}\left(\frac{2 x}{3}ight)ight](32)^{-1 / 5}\left(1+\frac{5}{32} xight)^{-1 / 5} \end{aligned}$ (Neglect higher powers of $x$ ) $=[1+x] 2^{-1}\left[1-\frac{1}{5}\left(\frac{5}{32}ight) xight]$ (Neglect higher powers of $x$ ) $\begin{aligned} & =\frac{1}{2}(1+x)\left(1-\frac{x}{32}ight) \ & =\frac{(1+x)(32-x)}{64}=\frac{32+31 x}{64} \end{aligned}$ (Neglect $x^2$ term)

If $x$ is numerically so small so that $x^2$ and higher powers of $x$ can be neglected, then…

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