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A text says For any positive integer n, Lim x approaching a $(x^n - a^n)/(x - a) = na^{n-1}$ Then it proved it using the quotient obtained by the division of two polynomials and obtained a polynomial and pluhged in x= a as polynomials are continuous to yield the result. Then it also said that the given theorem is also true if "N is any rational number and a is positive" why how can we prove this because then we no longer are dealing with polynomials ?

1 Answers 1

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By a change of variable:

Let $n=p/q$, then with $x=y^q$ and $a=b^q$,

$$\frac{x^{p/q} - a^{p/q}}{x - a}=\frac{y^p-b^p}{y^q-b^q}=\frac{\dfrac{y^p-b^p}{y-b}}{\dfrac{y^q-b^q}{y-b}},$$

which tends to

$$\frac{pb^{p-1}}{qb^{q-1}}=\frac{pa^{(p-1)/q}}{qa^{(q-1)/q}}=\frac pqa^{p/q-1}=na^{n-1}.$$

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    But what if p and q are negative integers n will still be rational but we cannot use previous result ? – 2017-02-14
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    @RaghavSingal: try it and you'll see. – 2017-02-14
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    Why does a has to be positive what if it is negative – 2017-02-14
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    @RaghavSingal: what is $(-1)^{1/2}$ ? – 2017-02-14
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    Okk I understand now – 2017-02-14