4. Let
f (x) =



x2−a2
x−a if x > a
x−a
x2−a2 if x < a
.
For what value(s) of a, if any does lim
x→a f (x) exist? For any value(s) that you find what
is the limit?

We are given the piecewise function:

\[
f(x) = \begin{cases} 
\frac{x^2 - a^2}{x - a} & \text{if } x > a \\
\frac{x - a}{x^2 - a^2} & \text{if } x < a 
\end{cases}
\]

### Step 1: Simplify the expression for \( x > a \)

For \( x > a \), we have:

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

The numerator \( x^2 - a^2 \) is a difference of squares, so we can factor it:

\[
f(x) = \frac{(x - a)(x + a)}{x - a}
\]

As long as \( x \neq a \), we can cancel out the \( x - a \) terms:

\[
f(x) = x + a
\]

Thus, for \( x > a \), the function simplifies to:

\[
f(x) = x + a
\]

### Step 2: Simplify the expression for \( x < a \)

For \( x < a \), we have:

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

Again, factor the denominator as a difference of squares:

\[
f(x) = \frac{x - a}{(x - a)(x + a)}
\]

For \( x \neq a \), we can cancel out the \( x - a \) terms:

\[
f(x) = \frac{1}{x + a}
\]

Thus, for \( x < a \), the function simplifies to:

\[
f(x) = \frac{1}{x + a}
\]

### Step 3: Evaluate the limit as \( x \to a \)

We now consider the left-hand limit and right-hand limit separately.

#### Right-hand limit (\( x \to a^+ \))

For \( x > a \), \( f(x) = x + a \). So as \( x \to a^+ \):

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

#### Left-hand limit (\( x \to a^- \))

For \( x < a \), \( f(x) = \frac{1}{x + a} \). So as \( x \to a^- \):

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

### Step 4: Set the limits equal to find \( a \)

For the limit \( \lim_{x \to a} f(x) \) to exist, the left-hand limit and right-hand limit must be equal. That is:

\[
2a = \frac{1}{2a}
\]

Multiply both sides by \( 2a \) to eliminate the fraction:

\[
(2a)(2a) = 1 \quad \Rightarrow \quad 4a^2 = 1 \quad \Rightarrow \quad a^2 = \frac{1}{4} \quad \Rightarrow \quad a = \pm \frac{1}{2}
\]

### Step 5: Evaluate the limit for \( a = \frac{1}{2} \) and \( a = -\frac{1}{2} \)

- If \( a = \frac{1}{2} \), the limit is \( \lim_{x \to \frac{1}{2}} f(x) = 2a = 1 \).
- If \( a = -\frac{1}{2} \), the limit is \( \lim_{x \to -\frac{1}{2}} f(x) = 2a = -1 \).

Thus, the limit exists for \( a = \pm \frac{1}{2} \), and the corresponding limits are:

- \( a = \frac{1}{2} \): \( \lim_{x \to a} f(x) = 1 \)
- \( a = -\frac{1}{2} \): \( \lim_{x \to a} f(x) = -1 \)

### Conclusion:

The limit exists for \( a = \pm \frac{1}{2} \), and the limit is:

- \( \lim_{x \to \frac{1}{2}} f(x) = 1 \)
- \( \lim_{x \to -\frac{1}{2}} f(x) = -1 \)

---

Would you like to dive into more details, or do you have any questions?

Here are five related questions to explore further:
1. How would the solution change if the function was not piecewise?
2. What happens to the limit if we modify the power of \(x\) in the expressions?
3. Can this type of function be extended to higher dimensions?
4. What is the significance of the difference of squares in simplifying limits?
5. How does the left-hand and right-hand limit analysis apply to other functions?

**Tip:** Always check both the left-hand and right-hand limits to ensure continuity at a point.

Let f(x) = { (x^2 - a^2) / (x - a) if x > a, (x - a) / (x^2 - a^2) if x < a }. For what value(s) of a, if any, does lim x→a f(x) exist? For any value(s) that you find, what is the limit?

Explore how to determine the values of 'a' for which the limit of a piecewise function exists. This solution simplifies expressions using algebraic techniques like factoring the difference of squares and evaluates left-hand and right-hand limits for continuity. Suitable for advanced high school mathematics students.

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