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The problem asks you to use the formal definition of a limit to prove that:

(a) \( \lim_{x \to 2} (x^2 - 3) = 1 \)

(b) \( \lim_{x \to 2} (2^x - 3) = 1 \)

The formal definition of a limit states:

For a function \( f(x) \), \( \lim_{x \to a} f(x) = L \) means that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that \( 0 < |x - a| < \delta \) implies \( |f(x) - L| < \epsilon \).

### Let's proceed with each part:

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### (a) Prove \( \lim_{x \to 2} (x^2 - 3) = 1 \) using the formal definition.

1. **Given function:** \( f(x) = x^2 - 3 \), and we want to show that as \( x \to 2 \), \( f(x) \to 1 \).

2. **Formal Definition:** We need to find a \( \delta > 0 \) such that if \( 0 < |x - 2| < \delta \), then \( |(x^2 - 3) - 1| < \epsilon \).

   Simplify:
   \[
   |(x^2 - 3) - 1| = |x^2 - 4| = |(x - 2)(x + 2)|
   \]

3. We now need \( |(x - 2)(x + 2)| < \epsilon \). To control \( |x + 2| \), note that for \( x \) near 2, we can assume \( |x - 2| < 1 \), which implies \( 1 < x < 3 \). Thus, \( |x + 2| < 5 \).

4. Now, we have:
   \[
   |(x - 2)(x + 2)| < 5|x - 2|
   \]

   To ensure \( 5|x - 2| < \epsilon \), set \( |x - 2| < \frac{\epsilon}{5} \).

5. Therefore, for \( \epsilon > 0 \), choose \( \delta = \frac{\epsilon}{5} \). This guarantees that \( |(x^2 - 3) - 1| < \epsilon \), which completes the proof.

---

### (b) Prove \( \lim_{x \to 2} (2^x - 3) = 1 \) using the formal definition.

1. **Given function:** \( f(x) = 2^x - 3 \), and we want to show that as \( x \to 2 \), \( f(x) \to 1 \).

2. **Formal Definition:** We need to find a \( \delta > 0 \) such that if \( 0 < |x - 2| < \delta \), then \( |(2^x - 3) - 1| < \epsilon \).

   Simplify:
   \[
   |(2^x - 3) - 1| = |2^x - 4|
   \]

3. We now need \( |2^x - 4| < \epsilon \). To control this, we can use the fact that \( 2^x \) is continuous and differentiable. The derivative \( \frac{d}{dx}(2^x) = 2^x \ln(2) \).

4. Using a linear approximation around \( x = 2 \) (since we are close to \( x = 2 \), the linear approximation works):
   \[
   2^x \approx 4 + (x - 2) \cdot 4 \ln 2
   \]
   Thus, we have:
   \[
   |2^x - 4| \approx 4 \ln 2 |x - 2|
   \]

5. To ensure \( 4 \ln 2 |x - 2| < \epsilon \), set \( |x - 2| < \frac{\epsilon}{4 \ln 2} \).

6. Therefore, for \( \epsilon > 0 \), choose \( \delta = \frac{\epsilon}{4 \ln 2} \). This guarantees that \( |(2^x - 3) - 1| < \epsilon \), completing the proof.

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Do you have any questions or want further clarifications on these steps?

Here are 5 related questions you might want to explore:
1. How does the epsilon-delta definition of a limit work for discontinuous functions?
2. How can we generalize the process of finding delta for more complex functions?
3. What are the implications of the continuity of exponential functions in limit proofs?
4. Can you explain the difference between one-sided limits and two-sided limits using epsilon-delta?
5. How can limits help in understanding instantaneous rates of change in calculus?

**Tip:** When using the epsilon-delta definition, always ensure to control every term in your inequality, especially when dealing with products or compositions of functions.

Using the formal definition of a limit to show that (a) lim_{x -> 2} (x^2 - 3) = 1 and (b) lim_{x -> 2} (2^x - 3) = 1.

Learn how to prove limits using the epsilon-delta definition for quadratic and exponential functions. This detailed step-by-step solution shows the process for (a) lim_{x -> 2} (x^2 - 3) = 1 and (b) lim_{x -> 2} (2^x - 3) = 1. Suitable for university-level calculus students.

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