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To solve the problem, I'll analyze each part step-by-step. The function \( f(x) \) is defined as follows:

\[
f(x) = \begin{cases} 
2 - 2x & \text{if } x < 2, \, x \neq 1 \\
\frac{-1}{x - 4} & \text{if } x > 2, \, x \neq 7 
\end{cases}
\]

Let's address each limit and the continuity question:

### a. \(\lim_{x \to 1} f(x)\)

Since \( x = 1 \) falls under the condition \( x < 2 \), we use the first piece of the function: \( f(x) = 2 - 2x \).

\[
\lim_{x \to 1} (2 - 2x) = 2 - 2(1) = 0
\]

Thus, \(\lim_{x \to 1} f(x) = 0\).

### b. \(\lim_{x \to 2^-} f(x)\) 

Approaching \( x = 2 \) from the left, we use the first piece of the function \( f(x) = 2 - 2x \):

\[
\lim_{x \to 2^-} (2 - 2x) = 2 - 2(2) = -2
\]

So, \(\lim_{x \to 2^-} f(x) = -2\).

### c. \(\lim_{x \to 2^+} f(x)\)

Approaching \( x = 2 \) from the right, we use the second piece of the function \( f(x) = \frac{-1}{x - 4} \):

\[
\lim_{x \to 2^+} \frac{-1}{x - 4} = \frac{-1}{2 - 4} = \frac{-1}{-2} = \frac{1}{2}
\]

Thus, \(\lim_{x \to 2^+} f(x) = \frac{1}{2}\).

### d. \(\lim_{x \to 7} f(x)\)

Since \( x = 7 \) falls under \( x > 2 \), we use the second piece: \( f(x) = \frac{-1}{x - 4} \):

\[
\lim_{x \to 7} \frac{-1}{x - 4} = \frac{-1}{7 - 4} = \frac{-1}{3}
\]

Therefore, \(\lim_{x \to 7} f(x) = \frac{-1}{3}\).

### e. Is \( f \) continuous at \( x = 7 \)? If not, is this reparable?

To check continuity at \( x = 7 \), the function must satisfy:
1. \( f(7) \) is defined,
2. \(\lim_{x \to 7} f(x)\) exists,
3. \(\lim_{x \to 7} f(x) = f(7)\).

The function \( f \) is not defined at \( x = 7 \), so \( f(7) \) does not exist. Therefore, \( f \) is not continuous at \( x = 7 \). 

**Reparability**: The discontinuity is removable. If we define \( f(7) = \frac{-1}{3} \), then the function would become continuous at \( x = 7 \).

### Summary of Results:
1. \(\lim_{x \to 1} f(x) = 0\)
2. \(\lim_{x \to 2^-} f(x) = -2\)
3. \(\lim_{x \to 2^+} f(x) = \frac{1}{2}\)
4. \(\lim_{x \to 7} f(x) = \frac{-1}{3}\)
5. \( f \) is not continuous at \( x = 7 \), but the discontinuity is reparable by defining \( f(7) = \frac{-1}{3} \).

Would you like further details or have any questions?

Here are some follow-up questions to consider:
1. What does it mean for a function to be piecewise continuous?
2. How would you determine if a limit exists when \( x \) approaches a specific point?
3. Can a function have a limit at a point where it is not defined?
4. What is the difference between a removable and a non-removable discontinuity?
5. How can a function be modified to make it continuous at a point of removable discontinuity?

**Tip**: When determining continuity at a point, always check if the limit from the left matches the limit from the right, and if both equal the function's value at that point.

Find the requested limits for the function f(x) defined as: f(x) = 2 - 2x if x < 2, x ≠ 1; f(x) = -1 / (x - 4) if x > 2, x ≠ 7. Also, check if f is continuous at x = 7 and if any discontinuity is reparable.

Solve the limits of a piecewise function f(x) with conditions at x = 1, x = 2, and x = 7. Determine the continuity of the function at x = 7 and whether the discontinuity can be repaired. Step-by-step calculus solution covering limits and continuity.

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