Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 500: 60

Answer

$\lim\limits_{x \to \infty}$ (1+ a/x)^(bx) = e^{-b}

Work Step by Step

$ \lim\limits_{x \to \infty} (1+ \frac{a}{x})^{bx}$ let $y = (1+ \frac{a}{x})^{bx}$ $ \ln(y) = \ln(1+ \frac{a}{x})^{bx}$ $ \lim\limits_{x \to \infty} (1+ \frac{a}{x})^{bx} = \lim\limits_{x \to \infty} (e^{\ln(1+ \frac{a}{x})^{bx}}) = \lim\limits_{x \to \infty} (e^{bx\ln(1+ \frac{a}{x})})$ Next: let $c = {bx\ln(1+ \frac{a}{x})}$ $\lim\limits_{x \to \infty} c =\frac{ {ln(1+ \frac{a}{x})}}{\frac{1}{bx}}$ which evaluates to $\frac{0}{0}$ so L'Hospital's Rule may be used giving : $\lim\limits_{x \to \infty} \frac{ {\frac{1}{1+ \frac{a}{x}}(\frac{-a}{x^{2}})}}{\frac{-1}{bx^{2}}}$ by cancelling out the $x^{2}$ on the topmost fraction and the bottom most fraction we get: $\lim\limits_{x \to \infty}\frac{-b}{1+\frac{a}{x}} = -b$ Plugging this into $e^{\ln(1+ \frac{a}{x})^{bx}}$ we get the final answer of: $e^{-b}$
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