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We are given a piecewise function \( f(x) \), and we are asked to determine if there exist real values for \( a \) and \( b \) such that the limits \( \lim_{x \to 0} f(x) \) and \( \lim_{x \to 3} f(x) \) exist.

The function is defined as:
\[
f(x) =
\begin{cases}
2x + a & \text{if } x < 0 \\
\frac{4 - \sqrt{4x + 4}}{x^2 - 2x - 3} & \text{if } x \geq 0 \text{ and } x \neq 3 \\
b & \text{if } x = 3
\end{cases}
\]

### Step 1: Analyze the limit as \( x \to 0 \)

We need to calculate:
\[
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (2x + a) = a
\]
and
\[
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{4 - \sqrt{4x + 4}}{x^2 - 2x - 3}
\]

First, simplify the expression for \( x \geq 0 \):
\[
4 - \sqrt{4x + 4} = 4 - 2\sqrt{x + 1}
\]
Thus, the expression becomes:
\[
f(x) = \frac{4 - 2\sqrt{x + 1}}{(x - 3)(x + 1)}
\]

Now, compute the limit:
\[
\lim_{x \to 0^+} \frac{4 - 2\sqrt{x + 1}}{(x - 3)(x + 1)}
\]
For \( x \to 0 \), this expression evaluates as:
\[
\frac{4 - 2\sqrt{1}}{(-3)(1)} = \frac{4 - 2}{-3} = \frac{2}{-3} = -\frac{2}{3}
\]

### Condition for the limit at \( x = 0 \) to exist:
\[
a = -\frac{2}{3}
\]

### Step 2: Analyze the limit as \( x \to 3 \)

Next, we evaluate the limits around \( x = 3 \):
\[
\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{4 - 2\sqrt{x + 1}}{(x - 3)(x + 1)}
\]
\[
\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{4 - 2\sqrt{x + 1}}{(x - 3)(x + 1)}
\]

Both limits involve a 0/0 indeterminate form, so we apply L'Hopital's rule:
1. Differentiate the numerator:
\[
\frac{d}{dx}[4 - 2\sqrt{x + 1}] = \frac{-1}{\sqrt{x + 1}}
\]
2. Differentiate the denominator:
\[
\frac{d}{dx}[(x - 3)(x + 1)] = (x - 3) + (x + 1) = 2x - 2
\]

Now apply L'Hopital's rule at \( x = 3 \):
\[
\lim_{x \to 3} \frac{\frac{-1}{\sqrt{x + 1}}}{2x - 2} = \frac{-1/\sqrt{4}}{6 - 2} = \frac{-1/2}{4} = -\frac{1}{8}
\]

### Condition for the limit at \( x = 3 \) to exist:
\[
b = -\frac{1}{8}
\]

### Conclusion:
The values of \( a \) and \( b \) that ensure the existence of the limits are:
\[
a = -\frac{2}{3}, \quad b = -\frac{1}{8}
\]

Would you like further details or have any questions?

---

Here are 5 related questions to deepen your understanding:
1. What is the importance of ensuring the limits match from both sides at a point of discontinuity?
2. How does L'Hopital's Rule help in resolving indeterminate forms like 0/0?
3. What would happen if the function had different expressions for \( x = 3 \) on both sides?
4. Can you find a case where the piecewise function has a jump discontinuity?
5. How do you compute higher-order limits at points of discontinuity?

**Tip:** Always check the behavior of the function from both sides when evaluating limits, especially for piecewise functions.

Determine if there exist values of a, b ∈ ℝ, such that the limits lim(x → 0) f(x) and lim(x → 3) f(x) exist for the given piecewise function.

This problem explores limits and continuity in a piecewise function, where real values for a and b must be found to ensure that the limits lim(x → 0) f(x) and lim(x → 3) f(x) exist. The solution involves simplifying expressions and applying L'Hopital's Rule to handle indeterminate forms.

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