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The problem asks to find the value of the constant \( a \) such that the limit of the function \( f(x) \) exists as \( x \) approaches 1. The function is defined piecewise:

\[
f(x) = 
\begin{cases}
ax^2 + 2x, & \text{if } x \leq 1 \\
\frac{x^2 - 1}{x - 1}, & \text{if } x > 1
\end{cases}
\]

To ensure that the limit exists at \( x = 1 \), we need to make sure the left-hand limit (\( \lim_{x \to 1^-} f(x) \)) and the right-hand limit (\( \lim_{x \to 1^+} f(x) \)) are equal, and that they also match \( f(1) \).

**Step 1: Calculate the left-hand limit (\( \lim_{x \to 1^-} f(x) \))**

Since \( x \leq 1 \), we use the first piece of the function:

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

**Step 2: Calculate the right-hand limit (\( \lim_{x \to 1^+} f(x) \))**

For \( x > 1 \), the function simplifies to:

\[
f(x) = \frac{x^2 - 1}{x - 1}
\]

This expression can be simplified as follows:

\[
f(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \text{ for } x \neq 1
\]

Thus:

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

**Step 3: Equating limits to find \( a \)**

For the limit to exist at \( x = 1 \), the left-hand and right-hand limits must be equal:

\[
a + 2 = 2
\]

Solving for \( a \):

\[
a = 0
\]

Therefore, the value of \( a \) that makes the limit exist as \( x \to 1 \) is \( \boxed{0} \).

Would you like more details on any of these steps, or do you have any other questions?

Here are some related questions:
1. What happens if we change the function's definition for \( x \leq 1 \)?
2. How do we handle continuity if the function has multiple points of interest?
3. What if the limit involves infinity? How do we handle such cases?
4. Can we have a limit if the function is undefined at the point of interest?
5. How do we approach limits involving trigonometric or exponential functions?

**Tip:** When working with limits, always check if simplifying expressions can help, especially when dealing with piecewise functions.

Find the value(s) of the constant a such that lim(x→1) f(x) exists, where f(x) is defined piecewise: f(x) = {ax^2 + 2x for x ≤ 1, (x^2 - 1)/(x - 1) for x > 1}.

This problem involves finding the constant a such that the limit of a piecewise function exists as x approaches 1. Learn how to handle limits for both left-hand and right-hand functions and ensure continuity.

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