Chapter 8.7, Problem 80E

To determine

Error in applying L’HospitaL’s rule in the limit

$\underset{x\to 0}{\lim }\frac{e^{2x}-1}{e^x}=\underset{x\to 0}{\lim }\frac{2e^{2x}}{e^x}=\underset{x\to 0}{\lim }2e^x=2$