If x is nearly equal to 1 then value of axb−bxaxb−xa is nearly equal to

Let substitute x=1+h in axb−bxaxb−xa=a1+hb-b(1+h)a(1+h)b-(1+h)a=a(1+bh)-b(1+ah)(1+bh)-(1+ah)=a-bh(b-a)=-1h ⇒axb−bxaxb−xa=-1h=11-x

If x is nearly equal to 1 then value of axb−bxaxb−xa is nearly equal to

Let substitute x=1+h in axb−bxaxb−xa=a1+hb-b(1+h)a(1+h)b-(1+h)a=a(1+bh)-b(1+ah)(1+bh)-(1+ah)=a-bh(b-a)=-1h ⇒axb−bxaxb−xa=-1h=11-x

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If x is nearly equal to 1 then value of $\frac{{ax}^{b} - {bx}^{a}}{x^{b} - x^{a}}$ is nearly equal to

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$\frac{2}{1 - x}$

$\frac{2}{1 + x}$

$\frac{1}{1 - x}$

$\frac{1}{1 + x}$

answer is B.

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Detailed Solution

Let $substitute x = 1 + h in \frac{{ax}^{b} - {bx}^{a}}{x^{b} - x^{a}} = \frac{a {(1 + h)}^{b} - b {(1 + h)}^{a}}{{(1 + h)}^{b} - {(1 + h)}^{a}} = \frac{a (1 + bh) - b (1 + ah)}{(1 + bh) - (1 + ah)} = \frac{a - b}{h (b - a)} = \frac{- 1}{h} \Rightarrow \frac{{ax}^{b} - {bx}^{a}}{x^{b} - x^{a}} = \frac{- 1}{h} = \frac{1}{1 - x}$