Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 56

Answer

The solution is $$\lim_{x\to 0}(1+4x)^{3/x}=e^{12}.$$

Work Step by Step

We will first transform the expression under the limit using the fact that the natural exponential function and the natural logarithmic function are inverse to each other i.e. that $a=e^{\ln a}$. This means that also $(1+4x)^{3/x}=e^{\ln(1+4x)^{3/x}}=e^{\frac{3}{x}\ln(1+4x)}$, where we also used the logarithmic rule that $\ln b^a=a\ln b$. Because the exponential function is continuous it can exchange places with the limit so we have $$L=\lim_{x\to 0}(1+4x)^{3/x}=\lim_{x\to0}e^{\frac{3}{x}\ln(1+4x)}=e^{3\lim_{x\to0}\frac{\ln(1+4x)}{x}}=e^{3l},$$ where we have denoted $l=\lim_{x\to0}\frac{\ln(1+4x)}{x}$. Let us now calculate this limit. "LR" will stand for "Apply L'Hopital's rule". $$l=\lim_{x\to0}\frac{\ln(1+4x)}{x}=\left[\frac{‚\ln(1+4\cdot 0)}{0}\right]=\left[\frac{\ln 1}{0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(\ln(1+4x))'}{(x)'}=\lim_{x\to0}\frac{\frac{1}{1+4x}(1+4x)'}{1}=\lim_{x\to0}\frac{4}{1+4x}=\frac{4}{1+4\cdot0}=4.$$ Putting this into the initial limit we get $$L=e^{3l}=e^{3\cdot4}=e^{12}.$$

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