The value of limx→0(1x+2x+3x+...+nxn)a/x is

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The value of

lim x \to 0 {(\frac{1^{x} + 2^{x} + 3^{x} + . . . + n^{x}}{n})}^{a / x}

is

Options:

A . (2n!)an

B . (n!)2an

C . (n!)an

D . 0

Answer: Option C
:
C

lim x \to 0 {(\frac{1^{x} + 2^{x} + 3^{x} + . . . + n^{x}}{n})}^{a / x} (1^{\infty} f o r m)

= e^{lim x \to 0} (\frac{1^{x} + 2^{x} + 3^{x} + . . . + n^{x}}{n} - 1) . \frac{a}{n}

= e^{lim x \to 0} (\frac{1^{x} + 2^{x} + 3^{x} + . . . + n^{x} - n}{n}) . \frac{a}{n}

= e^{lim x \to 0} (\frac{1^{x} + 2^{x} + 3^{x} + . . . + n^{x} - n}{x}) . \frac{a}{n}

= e^{lim x \to 0} {\frac{1^{x} - 1}{x} + \frac{2^{x} - 1}{x} + . . . + \frac{n^{x} - 1}{x}} \frac{a}{n}

We know

lim x \to 0 {\frac{a^{x} - 1}{x}} = a

So,

= e^{(log 1 + log 2 + log 3 + . . . log n) \frac{a}{n}}

e^{a / n} (log (1.2.3... n))

=

e^{({log}_{e} n!)^{a / n}} = (n!)^{a / n}

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