Proving Continuity at a Point: Techniques and Strategies

📚 Definition of Continuity at a Point

In calculus, a function $f(x)$ is said to be continuous at a point $x = a$ if the following three conditions are met:

• 🔍 $f(a)$ is defined (i.e., $a$ is in the domain of $f$).
• 📈 $\lim_{x \to a} f(x)$ exists. This means that both the left-hand limit and the right-hand limit exist and are equal.
• 🤝 $\lim_{x \to a} f(x) = f(a)$. The limit of the function as $x$ approaches $a$ is equal to the function's value at $a$.

📜 Historical Background

The concept of continuity, while seemingly intuitive, took mathematicians a while to formalize rigorously. Early notions were based on the idea of a function being "unbroken" or "smooth." Mathematicians like Cauchy and Weierstrass in the 19th century were instrumental in developing the precise epsilon-delta definition, providing a solid foundation for calculus and analysis.

• 🕰️ Early mathematicians relied on intuition regarding continuous curves.
• 👤 Cauchy and Weierstrass formalized the definition using limits.
• 💡 The epsilon-delta definition provided a rigorous approach to continuity.

🔑 Key Principles for Proving Continuity

To prove that a function is continuous at a point, you must verify all three conditions mentioned above. Here are the techniques:

• 📝 Evaluate $f(a)$: Make sure the function is defined at $x=a$. If it's undefined (e.g., division by zero), the function is discontinuous at that point.
• ⬅️➡️ Compute the Left-Hand Limit and Right-Hand Limit: The left-hand limit is denoted as $\lim_{x \to a^-} f(x)$ and the right-hand limit as $\lim_{x \to a^+} f(x)$. Both must exist and be equal for the limit to exist.
• ⚖️ Compare the Limit and Function Value: After finding the limit, ensure that it equals $f(a)$. If $\lim_{x \to a} f(x) \neq f(a)$, the function is discontinuous at $x = a$.

🌍 Real-world Examples

Let's examine some functions and determine their continuity at specified points.

1. Example 1: $f(x) = x^2 + 2x + 1$ at $x = 2$

   • ✅ $f(2) = (2)^2 + 2(2) + 1 = 4 + 4 + 1 = 9$. The function is defined at $x = 2$.
   • 📈 $\lim_{x \to 2} (x^2 + 2x + 1) = (2)^2 + 2(2) + 1 = 9$. The limit exists.
   • 🤝 $\lim_{x \to 2} f(x) = f(2) = 9$. The limit equals the function value. Therefore, $f(x)$ is continuous at $x = 2$.

2. Example 2: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$

   • ❌ $f(2) = \frac{(2)^2 - 4}{2 - 2} = \frac{0}{0}$, which is undefined. The function is undefined at $x=2$, so it is discontinuous there. Note: Even though the function can be simplified to $x+2$ for $x \neq 2$, it is still discontinuous at $x=2$ because the original function is undefined there.

3. Example 3: $f(x) = \begin{cases} x + 1, & x < 1 \\ 3 - x, & x \ge 1 \end{cases}$ at $x = 1$

   • ✅ $f(1) = 3 - 1 = 2$. The function is defined at $x = 1$.
   • ⬅️ $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 1) = 1 + 1 = 2$.
   • ➡️ $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3 - x) = 3 - 1 = 2$.
   • 🤝 Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = 2$, then $\lim_{x \to 1} f(x) = 2$. Also, $f(1) = 2$. Therefore, $\lim_{x \to 1} f(x) = f(1)$, and the function is continuous at $x = 1$.

🚀 Conclusion

Proving continuity at a point involves verifying that the function is defined at that point, the limit exists, and the limit equals the function's value. By carefully evaluating these conditions, we can rigorously determine whether a function is continuous at a given point. This concept is fundamental to understanding more advanced topics in calculus and analysis.