IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

AAKRITI

your answer has got the term 'a' while the question doesn't .

what i have figured out is that the expression should be

[ ( 1^x + 2^x +................+n^x ) / n ] ^a/x

i am sure only then the answer (n!)^a/n can come

the given limit is of the form 1^infinity

so its = e^{lim [ (1^x + 2^x +.......+n^x) /(n) -1 ]a/x} ( lim x -->0 everywhere )

= e^{lim[ {(1^x-1) +(2^x-1)........+ (n^x-1)}/x ]a/n}

= e^{[ ln1 + ln2 +............+lnn ]a/n}

= e^{[ ln(n!) ]a/n}

= (n!)^(a/n)