We are asked to evaluate the limit:

$\lim_{{x \to 0}} \frac{4e^x - 4x - 4}{13x^2}$

Step 1: Check for Indeterminate Form

Substitute $x = 0$ directly into the expression:

$\frac{4e^0 - 4(0) - 4}{13(0)^2} = \frac{4 - 4}{0} = \frac{0}{0}$

Since the limit results in the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule, which allows us to differentiate the numerator and denominator until the limit is no longer indeterminate.

Step 2: Apply L'Hôpital's Rule

Differentiate the numerator and denominator separately.

• Numerator: The derivative of $4e^x - 4x - 4$ is: $\frac{d}{dx}(4e^x - 4x - 4) = 4e^x - 4$

• Denominator: The derivative of $13x^2$ is: $\frac{d}{dx}(13x^2) = 26x$

Thus, the limit becomes:

$\lim_{{x \to 0}} \frac{4e^x - 4}{26x}$

Step 3: Check the New Indeterminate Form

Substitute $x = 0$ again:

$\frac{4e^0 - 4}{26(0)} = \frac{4 - 4}{0} = \frac{0}{0}$

We still have the indeterminate form $\frac{0}{0}$, so we apply L'Hôpital's Rule again.

Step 4: Apply L'Hôpital's Rule Again

Differentiate the numerator and denominator again.

• Numerator: The derivative of $4e^x - 4$ is: $\frac{d}{dx}(4e^x - 4) = 4e^x$

• Denominator: The derivative of $26x$ is: $\frac{d}{dx}(26x) = 26$

Thus, the limit becomes:

$\lim_{{x \to 0}} \frac{4e^x}{26}$

Step 5: Evaluate the Final Limit

Substitute $x = 0$:

$\frac{4e^0}{26} = \frac{4 \times 1}{26} = \frac{4}{26} = \frac{2}{13}$

Conclusion:

The value of the limit is:

$\boxed{\frac{2}{13}}$

Would you like further details or have any other questions?

Here are five related questions to consider:

1. How do you apply L'Hôpital's Rule to more complex limits?
2. What are the conditions under which L'Hôpital's Rule can be used?
3. How would this limit change if the exponent in $e^x$ was higher, say $e^{2x}$?
4. What other techniques can you use to solve limits if L'Hôpital’s Rule is not applicable?
5. Can L'Hôpital’s Rule be used for limits approaching infinity?

Tip: Always check the indeterminate form before applying L'Hôpital's Rule, as it can only be applied to specific forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.