The expression provided in the image is:

$\lim_{x \to 0} \frac{1 - e^{2x}}{1 - e^x}$

To evaluate this limit, let's proceed by examining the behavior of the function as $x \to 0$. The most effective method here is likely L'Hôpital's Rule, given that substituting $x = 0$ initially gives us an indeterminate form $\frac{0}{0}$.

Step 1: Applying L'Hôpital's Rule

Since both the numerator and denominator approach 0 as $x \to 0$, we can differentiate the numerator and the denominator with respect to $x$:

1. Differentiate the numerator $1 - e^{2x}$: $\frac{d}{dx}(1 - e^{2x}) = -2e^{2x}$

2. Differentiate the denominator $1 - e^x$: $\frac{d}{dx}(1 - e^x) = -e^x$

Thus, the limit becomes: $\lim_{x \to 0} \frac{-2e^{2x}}{-e^x} = \lim_{x \to 0} \frac{2e^{2x}}{e^x}$

Step 2: Simplify the Expression

Now, simplify the expression by dividing $e^{2x}$ by $e^x$: $\lim_{x \to 0} \frac{2e^{2x}}{e^x} = \lim_{x \to 0} 2e^{x}$

Step 3: Evaluate the Limit

As $x \to 0$, $e^x \to 1$. So, we can substitute directly:

$2e^{x} \to 2 \cdot 1 = 2$

Final Answer

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

Would you like a more detailed explanation of any part of this process?

Related Questions

1. What is L'Hôpital's Rule and when can it be applied?
2. How does the exponential function behave as $x \to 0$?
3. Could this limit be evaluated using a Taylor series expansion instead of L'Hôpital's Rule?
4. Why does the initial substitution result in an indeterminate form?
5. How would this limit change if the exponent in $e^{2x}$ were different?

Tip

When evaluating limits that result in indeterminate forms like $\frac{0}{0}$, L'Hôpital's Rule is often useful, but verifying with simpler methods (like substitution or expansion) can sometimes yield quicker insights.