When $\mathrm{x}$ is so small that its square and its higher powers may be neglected, then the value of $\frac{\left(1+\frac{3}{4} x\right)^{-4} \sqrt{(3+x)}}{\sqrt{(3-x)^3}}$ is approximately equal to

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Aicharya · Accepted Answer

Given that $x^n \approx 0, n=2,3,4 \ldots$ Let $y=\frac{\left(1+\frac{3}{4} xight)^{-4} \sqrt{(3+x)}}{\sqrt{(3-x)^3}}$ $ \begin{aligned} & =\frac{\left(1+\frac{3}{4} xight)^{-4} \sqrt{3}\left(1+\frac{1}{3} xight)^{1 / 2}}{(3)^{3 / 2}\left(1-\frac{1}{3} xight)^{3 / 2}} \ & \Rightarrow y=3^{1 / 2-3 / 2}\left(1+\frac{3}{4} xight)^{-4}(1+3 x)^{1 / 2}\left(1-\frac{1}{3} xight)^{-3 / 2} \ & \Rightarrow y=\frac{1}{3}\left[1+(-4)\left(\frac{3}{4} xight)+\frac{(-4)(-5)}{2 !}\left(\frac{3}{4} xight)^2+\ldotsight] \ & {\left[1+\left(-\frac{3}{2}ight)\left(\frac{-1}{3} xight)+\frac{(-3 / 2)(-3 / 2-1)}{2 !}\left(-\frac{1}{3} xight)^2+\ldotsight]} \end{aligned} $ Now we will neglect term with $x^2$ and greater powers. i.e. $x^2=x^3=x^4=\ldots=x^n \approx 0$. $ \begin{aligned} & \Rightarrow y=\frac{1}{3}[1-3 x+0+\ldots]\left[1+\frac{1}{6} x-0+\ldotsight] \ & \Rightarrow y=\frac{1}{3}(1-3 x)\left(1+\frac{1}{6} xight)\left(1+\frac{1}{2} xight) \ & \Rightarrow y=\frac{1}{3}\left(1+\frac{1}{6} x-3 x-\frac{3 x^2}{6}ight)\left(1+\frac{1}{2} xight) \ & \Rightarrow y=\frac{1}{3}\left(1+\frac{1}{2} x-\frac{17}{6} x-\frac{17}{12} x^2ight) \ & \Rightarrow y=\frac{1}{3}\left(1-\frac{14}{6} xight)=\frac{1}{3}\left(1-\frac{7}{3} xight) \Rightarrow y=\frac{1}{3}-\frac{7}{9} x \end{aligned} $

When $\mathrm{x}$ is so small that its square and its higher powers may be neglected, then the value of…

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