First slide

Introduction to limits

Question

$lim_{x \to 0} {\{\frac{1^{x} + 2^{x} + 3^{x} + \dots + n^{x}}{n}\}}^{1 / x}$ is equal to

Moderate

A

$(n!)^{n}$
B

$(n!)^{1 / n}$
C

$n!$
D

$\ln (n!)$

Solution

We have,
$\begin{array}{l} lim_{x \to 0} {(\frac{1^{x} + 2^{x} + \dots + n^{x}}{n})}^{1 / x} \\ = lim_{x \to 0} {\{1 + \frac{1^{x} - 1}{n} + \frac{2^{x} - 1}{n} + \dots + \frac{n^{x} - 1}{n}\}}^{1 / x} \\ = e^{lim_{x \to 0} \frac{1}{n} \{\frac{1^{x} - 1}{x} + \frac{2^{x} - 1}{x} + \dots + \frac{n^{x} - 1}{x}\}} \\ = e^{\frac{1}{n} [\log 1 + \log 2 + \dots + \log n]} \end{array}$
$= e^{\frac{1}{n} (\log n!)} = e^{\log (n!)^{\frac{1}{n}}} = (n!)^{\frac{1}{n}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App