Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.5 L'Hopital's Rule; Indeterminate Forms - Exercises Set 6.5 - Page 448: 17

Answer

$0$

Work Step by Step

Our aim is to evaluate the limit for $\lim\limits_{x \to +\infty} \dfrac{ x^{100}}{e^x}$. But $\lim\limits_{x \to +\infty} \dfrac{ x^{100}}{e^x}=\dfrac{\infty}{\infty}$ We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{\infty}{\infty}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{A(x) }{B(x)}=\lim\limits_{x \to a}\dfrac{A'(x)}{B'(x)}$ where, $a$ can be any real number, infinity or negative infinity. $\lim\limits_{x \to +\infty} \dfrac{ x^{100}}{e^x}=\lim\limits_{x \to +\infty} \dfrac{ 100x^{99}}{e^x}=\dfrac{\infty}{\infty}$ Again apply L'Hopital's rule : $\lim\limits_{x \to +\infty} \dfrac{ 100x^{99}}{e^x}=\lim\limits_{x \to + \infty} \dfrac{(100)(98)x^{98}}{e^x}=\lim\limits_{x \to +\infty} \dfrac{ 100 !}{e^x}=100! \times \lim\limits_{x \to +\infty} e^{-x}=100 ! \times 0=0$
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