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To solve the limit problem, we need to evaluate:

\[
\lim_{x \to 1} \left( \frac{7x}{x - 1} - \frac{7}{\ln(x)} \right)
\]

### Step 1: Simplify the Expression
Let's examine each term individually as \( x \to 1 \):

1. **The first term** \( \frac{7x}{x - 1} \): As \( x \to 1 \), the numerator approaches \( 7 \) while the denominator approaches \( 0 \), so this term tends toward \( \frac{7}{0} \), suggesting an infinite behavior. To make this precise, we can use L'Hospital's Rule.

2. **The second term** \( \frac{7}{\ln(x)} \): As \( x \to 1 \), \( \ln(x) \to 0 \), and the fraction \( \frac{7}{\ln(x)} \) diverges. Again, L'Hospital's Rule can help.

### Step 2: Apply L'Hospital's Rule
Since both terms lead to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) forms, we apply L'Hospital's Rule to each term separately.

#### First Term: \( \frac{7x}{x - 1} \)
Using L'Hospital's Rule:
\[
\frac{d}{dx}(7x) = 7 \quad \text{and} \quad \frac{d}{dx}(x - 1) = 1,
\]
so
\[
\lim_{x \to 1} \frac{7x}{x - 1} = \lim_{x \to 1} \frac{7}{1} = 7.
\]

#### Second Term: \( \frac{7}{\ln(x)} \)
Using L'Hospital's Rule:
\[
\frac{d}{dx}(7) = 0 \quad \text{and} \quad \frac{d}{dx}(\ln(x)) = \frac{1}{x},
\]
so
\[
\lim_{x \to 1} \frac{7}{\ln(x)} = \lim_{x \to 1} \frac{0}{1/x} = \lim_{x \to 1} 0 \cdot x = 0.
\]

### Step 3: Combine Results
Now we have:
\[
\lim_{x \to 1} \left( \frac{7x}{x - 1} - \frac{7}{\ln(x)} \right) = 7 - 0 = 7.
\]

### Final Answer:
\[
\boxed{7}
\]

Would you like further explanation on any step?

### Follow-up Questions:
1. How is L'Hospital's Rule applied to other indeterminate forms?
2. What other indeterminate forms are there besides \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \)?
3. How can Taylor series be used in evaluating limits?
4. What is the difference between one-sided and two-sided limits?
5. How would the solution change if the limit approached a value other than \( x = 1 \)?

### Tip:
Always check if direct substitution or simplification can solve the limit before using L'Hospital’s Rule—it can save time and make the process simpler.

Find the limit: \( \lim_{x \to 1} \left( \frac{7x}{x - 1} - \frac{7}{\ln(x)} \right) \). Use l'Hôpital's Rule where appropriate. If there is a more elementary method, consider using it.

Learn how to evaluate the limit \( \lim_{x \to 1} \left( \frac{7x}{x - 1} - \frac{7}{\ln(x)} \right) \) using L'Hôpital's Rule. This problem covers key calculus concepts such as limits, indeterminate forms, and applying L'Hôpital's Rule. Suitable for undergraduate calculus students.

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