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Determiner la limite en utilisant la règle de l’hôpital

To determine the given limit:

\[
\lim_{x \to 2} \frac{e^x - e^{-2}}{x^2 + x - 2}
\]

we can start by substituting \(x = 2\) directly into the expression to see if it results in an indeterminate form (like \( \frac{0}{0} \)):

- In the numerator: \( e^2 - e^{-2} \).
- In the denominator: \( 2^2 + 2 - 2 = 4 + 2 - 2 = 4 \).

This leads to:

\[
\frac{e^2 - e^{-2}}{4} = \frac{\text{some non-zero value}}{4}.
\]

Since this is a finite value, the limit can be directly computed without using L'Hôpital's Rule. The value of the limit is:

\[
\lim_{x \to 2} \frac{e^x - e^{-2}}{x^2 + x - 2} = \frac{e^2 - e^{-2}}{4}.
\]

Would you like more details on the next steps or a breakdown of this calculation?

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Here are some questions for further exploration:
1. What is L'Hôpital's Rule and when is it applicable?
2. What happens if the numerator and denominator both approach zero?
3. How do we find limits involving exponential functions more generally?
4. Can this limit be evaluated using Taylor series expansion?
5. What other methods could be used to calculate limits like this?

**Tip:** Always substitute first to check if a direct limit can be found before applying L'Hôpital's Rule.

Déterminer la limite: lim(x → 2) [(e^x - e^(-2)) / (x^2 + x - 2)]

Learn how to determine the limit of (e^x - e^(-2)) / (x^2 + x - 2) as x approaches 2. This calculus problem involves checking for indeterminate forms and using L'Hôpital's Rule to find the solution. Suitable for students in Grades 11-12.

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