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

Step-by-step solution

1. The limit is of indeterminate form $0/0$, so apply L'Hôpital's rule.
2. Differentiate numerator: $(e^{2x}\cdot 2)-2=2e^{2x}-2$. Differentiate denominator: $2x$.
3. Now the limit becomes ${\lim }_{x\to 0}\frac{2e^{2x}-2}{2x}={\lim }_{x\to 0}\frac{e^{2x}-1}{x}$.
4. Again $0/0$, apply L'Hôpital's rule a second time.
5. Differentiate numerator: $2e^{2x}$. Differentiate denominator: $1$.
6. Now limit is ${\lim }_{x\to 0}2e^{2x}=2\cdot e^0=2$.
7. Thus the required limit equals $2$.

Concept tested: Limit using L'Hôpital's rule

Published 3 September 2026 · How this solution was prepared and checked