## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises: 56

#### Answer

$\lim\limits_{n \to \infty}(1+\frac{x}{n})^{n}=e^{x}$

#### Work Step by Step

Given: $\lim\limits_{n \to \infty}(1+\frac{x}{n})^{n}=e^{x};x>0$ From the definition of derivative as a limit, we get $f'(y) =\lim\limits_{h \to 0}\frac{ln(y+h)-lny}{h}$ Apply logarithmic property. $\lim\limits_{h \to 0}\frac{\frac{ln(y+h)}{y}}{h}=\frac{1}{y}$ Suppose $h=\frac{1}{n}$ and $y=\frac{1}{x}$ when ${h \to 0}$ then ${n \to \infty}$ Thus, $\lim\limits_{h \to 0}\frac{\frac{ln(1/x+1/n)}{1/x}}{1/n}=\frac{1}{1/x}$ $\lim\limits_{n \to \infty}(1+\frac{x}{n})^{n}=x$ Take exponent on both sides of above expression. $e^{\lim\limits_{n \to \infty}ln(1+\frac{x}{n})^{n}}=e^{x}$ Since, $e^{lnx}=x$ Hence, $\lim\limits_{n \to \infty}(1+\frac{x}{n})^{n}=e^{x}$

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