Discussion
The value of \( \lim_{x \to 0} \dfrac{e^{2x} - 1}{\tan 3x} \) is
Answer: B
Replace each factor by its leading term, \( e^{2x} - 1 \approx 2x \) and \( \tan 3x \approx 3x \):
\[ \lim_{x \to 0}\frac{e^{2x}-1}{\tan 3x} \;=\; \lim_{x \to 0}\frac{2x}{3x} \;=\; \frac{2}{3}. \]
L'Hopital's rule gives the same result, the derivatives being \( 2e^{2x} \) and \( 3\sec^{2}3x \), both equal to their coefficients at the origin
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