If $x$ is small, so that $x^2$ and higher powers can be neglected, then the approximate value for $\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}$ is

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Aicharya · Accepted Answer

Given expression is $ \begin{aligned} & E=\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}} \ &=\frac{\left(1+2 x+2 x^2+\ldotsight)(1+6 x+\ldots)}{(1+12 x+\ldots)} \ &=\frac{(1+2 x+6 x+\ldots)}{(1+12 x)} \ &=(1+8 x)(1+12 x)^{-1} \ &=(1+8 x)(1-12 x \ldots) \ &=(1+8 x-12 x \ldots) \ &=(1-4 x) \ & \quad(\because 	ext { neglecting the higher term }) \end{aligned} $ ( $\because$ neglecting the higher term)

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

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