To prove : lim\( _{x \to a} { x^n – a^n \over x – a } = na^{n-1} \) where n is a rational number

Proof:

Let  \( x = a + h  \)

Then as  \(x \to a \), we have  \(h \to 0 \)

Now, \( \lim_{x \to a} { x^n – a^n \over x – a } =  \lim_{h \to 0} { (a + h)^n – a^n \over { a + h – a } } \)

= \( \lim_{h \to 0} { a^n ( 1 + {h \over a} )^n – a^n \over h } \)      { Taking \( a^n \) out as a common factor }

= \( \lim_{h \to 0} { a^n [ ( 1 + {h \over a} )^n – 1 ] \over h } \)   

= \( \lim_{h \to 0} { a^n [   \{  1 + {n \over 1! }( { h \over a} )^1  + {n(n-1) \over 2! }( { h \over a} )^2 + {n(n-1)(n-2) \over 3! }( { h \over a} )^3 + … + {n(n-1)(n-2)…3.2.1 \over n! } ( { h \over a} )^n \}  – 1 ] \over h }  \)

= \( \lim_{h \to 0} { a^n [  {n \over 1! }( { h \over a} )^1  + {n(n-1) \over 2! }( { h \over a} )^2 + {n(n-1)(n-2) \over 3! }( { h \over a} )^3 + … + {n(n-1)(n-2)…3.2.1 \over n! } ( { h \over a} )^n ] \over h }  \)

= \( \lim_{h \to 0} { a^n.n.h [  {1 \over 1! } { 1 \over a^1} + {(n-1) \over 2! } { h^1 \over a^2} + {(n-1)(n-2) \over 3! } { h^2 \over a^3} + … + {(n-1)(n-2)…3.2.1 \over n! } { h^{n-1} \over a^n} ] \over h }  \)     { Taking n & h out as common factor }

= \( \lim_{h \to 0} { a^n.n.{h \over a} [  {1 \over 1! } + {(n-1) \over 2! } ({ h \over a})^1 + {(n-1)(n-2) \over 3! } ({ h \over a})^2 + … + {(n-1)(n-2)…3.2.1 \over n! } ({ h \over a})^{n-1} ] \over h }  \)

= \( \lim_{h \to 0} { na^{n-1} \, [  {1 \over 1! } + {(n-1) \over 2! } ({ h \over a})^1 + {(n-1)(n-2) \over 3! } ({ h \over a})^2 + … + {(n-1)(n-2)…3.2.1 \over n! } ({ h \over a})^{n-1} ] }  \)

= \( n a ^{n-1}\)