The value of lim x → 0 ( 1 x + 2 x + 3 x + . . . + n x n ) a / x is Options: A . &nbsp ( 2 n ! ) a n B . &nbsp ( n ! ) 2 a n C . &nbsp ( n ! ) a n D . &nbsp 0

Answer: Option C : C lim x → 0 ( 1 x + 2 x + 3 x + . . . + n x n ) a / x ( 1 ∞ f o r m ) = e lim x → 0 ( 1 x + 2 x + 3 x + . . . + n x n − 1 ) . a n = e lim x → 0 ( 1 x + 2 x + 3 x + . . . + n x − n n ) . a n = e lim x → 0 ( 1 x + 2 x + 3 x + . . . + n x − n x ) . a n = e lim x → 0 { 1 x − 1 x + 2 x − 1 x + . . . + n x − 1 x } a n We know lim x → 0 { a x − 1 x } = a So, = e ( log 1 + log 2 + log 3 + . . . log n ) a n e a / n ( log ( 1.2.3... n ) ) = e ( log e n ! ) a / n = ( n ! ) a / n

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

Discussion Forum : Limits Continuity And Differentiability

Question -

The value of $lim x \to 0 {(\frac{1^{x} + 2^{x} + 3^{x} + . . . + n^{x}}{n})}^{a / x}$ is

Options:

A .

(2 n!)^{\frac{a}{n}}

B .

(n!)^{\frac{2 a}{n}}

C .

(n!)^{\frac{a}{n}}

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}

Was this answer helpful ?

Next Question

Discussion Forum : Limits Continuity And Differentiability

Submit Your Solution hear: