Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Practice Exercises - Page 440: 108

Answer

= $1$

Work Step by Step

$f(x)$ = $\mathop {\lim }\limits_{x \to 0^{+} }$ $(1+\frac{3}{x})^{x}$ $\ln{f(x)}$ = $\mathop {\lim }\limits_{x \to 0^{+} }$ $\ln(1+\frac{3}{x})^{x}$ $\ln{f(x)}$ = $\mathop {\lim }\limits_{x \to 0^{+} }$ $x(\ln(1+3x^{-1}))$ $\ln{f(x)}$ = $\mathop {\lim }\limits_{x \to 0^{+} }$ $\frac{\ln(1+3x^{-1})}{x^{-1}}$ $\ln{f(x)}$ = $\mathop {\lim }\limits_{x \to 0^{+} }$ $\frac{\frac{1}{1+3x^{-1}}(-3x^{-2})}{-x^{-2}}$ $\ln{f(x)}$ = $\mathop {\lim }\limits_{x \to 0^{+} }$ $\frac{3}{1+3x^{-1}}$ $\ln{f(x)}$ = $\mathop {\lim }\limits_{x \to 0^{+} }$ $\frac{3x}{x+3}$ $\ln{f(x)}$ = $0$ ${f(x)}$ = $e^{0}$ ${f(x)}$ = $1$
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