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The problem asks to estimate the limit numerically for the following expression:

\[
\lim_{{x \to 0}} \frac{{6^x - 1}}{{x}}
\]

### Step 1: Apply Limit Definition
This limit is in the form of \(\frac{{0}}{{0}}\), so we can apply **L'Hôpital's Rule**, which states that if a limit results in the indeterminate form \(\frac{0}{0}\), we can differentiate the numerator and the denominator separately and then re-evaluate the limit.

The numerator is \( 6^x - 1 \), and the denominator is \( x \).

### Step 2: Differentiate the Numerator and Denominator
- The derivative of \( 6^x \) is \( 6^x \ln(6) \).
- The derivative of \( -1 \) is 0.
- The derivative of \( x \) is 1.

Thus, applying L'Hôpital's Rule:

\[
\lim_{{x \to 0}} \frac{{6^x - 1}}{{x}} = \lim_{{x \to 0}} \frac{{6^x \ln(6)}}{{1}} = 6^0 \ln(6) = \ln(6)
\]

### Step 3: Calculate Numerically
\[
\ln(6) \approx 1.791759
\]

Thus, the limit is approximately:

\[
\boxed{1.792}
\]

Would you like more details or have any questions about the process?

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Here are 5 related questions to deepen your understanding:

1. How does L'Hôpital's Rule work, and in what cases can it be applied?
2. What are some alternative methods to evaluate limits without using L'Hôpital's Rule?
3. Can the limit of \( \frac{6^x - 1}{x} \) be evaluated using a Taylor series expansion of \( 6^x \)?
4. What happens if we replace the base 6 in \( 6^x \) with another number?
5. Why is the derivative of \( 6^x \) equal to \( 6^x \ln(6) \)?

**Tip:** Always check if a limit produces an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) to decide if L'Hôpital's Rule is applicable.

Estimate the limit numerically or state that the limit does not exist: lim(x→0) (6^x - 1) / x

Learn how to estimate the limit lim(x→0) (6^x - 1) / x using L'Hôpital's Rule. This problem focuses on the application of limits, differentiation, and exponential functions. Ideal for students in Grades 11-12 or Calculus learners.

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