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

1 - 2 x

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1 - 3 x

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1 - 4 x

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1 - 5 x

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Solution

The correct option is C $1 - 4 x$
$\frac{(1 - 2 x)^{- 1} (1 - 3 x)^{- 2}}{(1 - 4 x)^{- 3}} = (1 + 2 x) (1 + 6 x) (1 - 4 x)^{3} = (1 + 2 x) (1 + 6 x) (1 - 12 x) = 1 - 4 x$
Since if $x$ is very small, then $(1 + x)^{n} = 1 + n x$

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