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The problem asks to determine the limit of the piecewise function \( f(x) \) as \( x \to 1 \).

The function is defined as:

\[
f(x) = 
\begin{cases} 
x + 1 & \text{if } x < 1 \\
-x^2 + 4x - 1 & \text{if } x \geq 1 
\end{cases}
\]

To solve the limit \( \lim_{x \to 1} f(x) \), we need to check the left-hand limit and the right-hand limit separately, then compare them.

### Left-hand limit (\( x \to 1^{-} \)):
When \( x \to 1 \) from the left (\( x < 1 \)), we use the part of the function \( f(x) = x + 1 \).

\[
\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} (x + 1) = 1 + 1 = 2
\]

### Right-hand limit (\( x \to 1^{+} \)):
When \( x \to 1 \) from the right (\( x \geq 1 \)), we use the part of the function \( f(x) = -x^2 + 4x - 1 \).

\[
\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} (-x^2 + 4x - 1) = -(1)^2 + 4(1) - 1 = -1 + 4 - 1 = 2
\]

### Conclusion:
Since both the left-hand limit and right-hand limit are equal to 2, the overall limit exists and is:

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

Would you like more details or have any further questions?

Here are 5 related questions to expand your understanding:

1. What is the general process to solve limits for piecewise functions?
2. How do you determine continuity at \( x = 1 \) for this function?
3. What happens if the left-hand and right-hand limits do not agree?
4. How would the limit change if the functions on either side of \( x = 1 \) were different?
5. Can you find the derivative of \( f(x) \) at \( x = 1 \) using limits?

**Tip:** Always check both the left-hand and right-hand limits when dealing with piecewise functions. The function's limit exists at a point if and only if both limits are equal.

Determine the limit of the piecewise function f(x) as x approaches 1.

Explore how to determine the limit of a piecewise function as x approaches 1. This solution walks through finding both the left-hand and right-hand limits to conclude whether the overall limit exists.

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