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We are asked to evaluate the limit:

\[
\lim_{x \to 0} \frac{12e^x - 12x - 12}{12x^2}
\]

### Step 1: Substituting \( x = 0 \)
If we directly substitute \( x = 0 \), we get:

\[
\frac{12e^0 - 12(0) - 12}{12(0)^2} = \frac{12(1) - 0 - 12}{0} = \frac{0}{0}
\]

This is an indeterminate form (\( \frac{0}{0} \)), so we can apply **L'Hôpital's Rule**.

### Step 2: Applying L'Hôpital's Rule
L'Hôpital's Rule states that if we have an indeterminate form of \( \frac{0}{0} \), we can differentiate the numerator and the denominator and then take the limit again. We need to find the derivatives of the numerator and denominator.

- **Numerator**: \( 12e^x - 12x - 12 \)
  
  The derivative is:
  \[
  \frac{d}{dx}(12e^x - 12x - 12) = 12e^x - 12
  \]

- **Denominator**: \( 12x^2 \)
  
  The derivative is:
  \[
  \frac{d}{dx}(12x^2) = 24x
  \]

### Step 3: Taking the limit again
Now, the new limit becomes:

\[
\lim_{x \to 0} \frac{12e^x - 12}{24x}
\]

Substitute \( x = 0 \) again:

\[
\frac{12e^0 - 12}{24(0)} = \frac{12(1) - 12}{0} = \frac{0}{0}
\]

This is still indeterminate, so we apply **L'Hôpital's Rule** again.

### Step 4: Applying L'Hôpital's Rule again
We differentiate the numerator and denominator once more.

- **Numerator**: \( 12e^x - 12 \)

  The derivative is:
  \[
  \frac{d}{dx}(12e^x - 12) = 12e^x
  \]

- **Denominator**: \( 24x \)

  The derivative is:
  \[
  \frac{d}{dx}(24x) = 24
  \]

### Step 5: Taking the final limit
Now, the new limit becomes:

\[
\lim_{x \to 0} \frac{12e^x}{24}
\]

Substitute \( x = 0 \):

\[
\frac{12e^0}{24} = \frac{12(1)}{24} = \frac{12}{24} = \frac{1}{2}
\]

### Conclusion:
The value of the limit is \( \frac{1}{2} \), so the limit exists and is neither infinity nor undefined.

Would you like more details on L'Hôpital's Rule or the differentiation process?

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Here are 5 related questions for further exploration:
1. How does L'Hôpital's Rule work for limits involving \(\infty/\infty\) forms?
2. Can L'Hôpital's Rule be used for limits involving higher-order indeterminacies?
3. What happens if the limit does not resolve after multiple applications of L'Hôpital's Rule?
4. How do we handle limits when L'Hôpital's Rule is not applicable?
5. What are the alternatives to L'Hôpital's Rule for finding limits?

**Tip**: Always verify that you're dealing with an indeterminate form (\(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)) before applying L'Hôpital's Rule!

Evaluate the limit using L'Hôpital's Rule: lim (x -> 0) [(12e^x - 12x - 12) / (12x^2)]

Learn how to evaluate the limit using L'Hôpital's Rule for the expression lim (x -> 0) [(12e^x - 12x - 12) / (12x^2)]. This problem involves applying derivatives multiple times to resolve an indeterminate form. Suitable for high school and college calculus students.

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