Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 42

Answer

The solution is $$\lim_{x\to\infty}\frac{\ln(3x+5e^x)}{\ln(7x+3e^{2x})}=\frac{1}{2}.$$

Work Step by Step

To solve this limit follow the steps below $$\lim_{x\to\infty}\frac{\ln(3x+5e^x)}{\ln(7x+3e^{2x})}=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(\ln(3x+5e^x))'}{(\ln(7x+3e^{2x}))'}=\lim_{x\to\infty}\frac{\frac{1}{3x+5e^x}(3x+5e^x)'}{\frac{1}{7x+3e^{2x}}(7x+3e^{2x})'}=\lim_{x\to\infty}\frac{\frac{3+5e^x}{3x+5e^x}}{\frac{7+3e^{2x}(2x)'}{7x+3e^{2x}}}=\lim_{x\to\infty}\frac{\frac{3+5e^x}{3x+5e^x}}{\frac{7+6e^{2x}}{7x+3e^{2x}}}=\lim_{x\to\infty}\frac{(3+5e^x)(7x+3e^{2x})}{(3x+5e^x)(7+6e^{2x})}=\lim_{x\to0}\frac{e^{3x}(3e^{-x}+5)(7(x/e^{2x})+3)}{e^{3x}(3(x/e^x)+5)(7e^{-2x}+6)}=\lim_{x\to0}\frac{(3e^{-x}+5)(7(x/e^{2x})+3)}{(3(x/e^x)+5)(7e^{-2x}+6)}.$$ Note that, since every exponential function grows faster as $x\to\infty$ than any power function we have that $x/e^{2x}\to0$ and $x/e^x\to0$. Putting this into our limit $$\lim_{x\to\infty}\frac{\ln(3x+5e^x)}{\ln(7x+3e^{2x})}=\left[\frac{(3e^{-\infty}+5)(7\cdot0+3)}{(3\cdot0+5)(7e^{-2\cdot\infty}+6)}\right]=\left[\frac{5\cdot3}{5\cdot6}\right]=\frac{1}{2}.$$
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