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

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Aicharya · Accepted Answer

Using binomial expansions and neglecting $x^2$ and higher powers, we get $ \begin{aligned} \frac{\left[2\left(1+\frac{x}{4}ight)^{\frac{1}{2}}+2\left(1-\frac{x}{8}ight)^{\frac{1}{3}}ight]}{\left(1-\frac{2 x}{3}ight)^{\frac{3}{2}}} \ =2\left[1+\frac{1}{2} \cdot \frac{x}{4}+1-\frac{1}{3} \cdot \frac{x}{8}ight]\left(1-\frac{2 x}{3}ight)^{-\frac{3}{2}} \ =2\left[2+\frac{x}{8}-\frac{x}{24}ight]\left[1-\frac{3}{2} \cdot\left(\frac{-2 x}{3}ight)ight] \ =2\left[2+\frac{2 x}{24}ight][1+x] \end{aligned} $ $ \begin{aligned} & =2\left[2+\frac{x}{12}ight][1+x]=2\left[2+2 x+\frac{x}{12}ight] \ & =2\left[2+\frac{25 x}{12}ight] \end{aligned} $ When, $x=\frac{6}{25}$ $ =2\left[2+\frac{25}{12} \cdot \frac{6}{25}ight]=2\left[2+\frac{1}{2}ight]=2 \cdot \frac{5}{2}=5 $

If $|x|$ is so small that $x^2$ and higher powers of $x$ may be neglected, then the approximate value of…

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