Worked Problems: Demonstrating Continuity on an Interval Formally

📚 Understanding Continuity on an Interval

Continuity on an interval means that a function is continuous at every point within that interval. This includes both open and closed intervals. Formally, a function $f(x)$ is continuous on an interval $[a, b]$ if it is continuous at every point $c$ in $(a, b)$ and if the one-sided limits exist and equal the function values at the endpoints $a$ and $b$. Let's dive into some examples!

📜 Historical Context

The formal definition of continuity evolved over centuries. Mathematicians like Cauchy and Weierstrass formalized the concept using limits, addressing earlier intuitive notions that lacked rigor. This formalization is crucial for advanced calculus and analysis.

✨ Key Principles of Continuity on an Interval

• 🔍 Definition of Continuity at a Point: A function $f(x)$ is continuous at a point $c$ if $\lim_{x \to c} f(x) = f(c)$. This means the limit exists, the function is defined at $c$, and the limit equals the function value.
• 💡 One-Sided Limits: For a function to be continuous on a closed interval $[a, b]$, the following must hold:
  • $\lim_{x \to a^+} f(x) = f(a)$ (right-hand limit at $a$)
  • $\lim_{x \to b^-} f(x) = f(b)$ (left-hand limit at $b$)
• 📝 Intermediate Value Theorem (IVT): If $f(x)$ is continuous on $[a, b]$ and $k$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in $(a, b)$ such that $f(c) = k$.
• 📈 Extreme Value Theorem (EVT): If $f(x)$ is continuous on $[a, b]$, then $f(x)$ must attain a maximum and a minimum value on $[a, b]$.

🔢 Worked Problems

Example 1: $f(x) = x^2$ on $[-1, 2]$

Prove that $f(x) = x^2$ is continuous on the closed interval $[-1, 2]$.

1. ✅ Continuity on the Open Interval $(-1, 2)$: $f(x) = x^2$ is a polynomial, and polynomials are continuous everywhere. Thus, $f(x)$ is continuous for all $x$ in $(-1, 2)$.
2. ➡️ Right-Hand Limit at $x = -1$: We need to show that $\lim_{x \to -1^+} f(x) = f(-1)$. $$\lim_{x \to -1^+} x^2 = (-1)^2 = 1$$ Since $f(-1) = (-1)^2 = 1$, the right-hand limit equals the function value at $x = -1$.
3. ⬅️ Left-Hand Limit at $x = 2$: We need to show that $\lim_{x \to 2^-} f(x) = f(2)$. $$\lim_{x \to 2^-} x^2 = (2)^2 = 4$$ Since $f(2) = (2)^2 = 4$, the left-hand limit equals the function value at $x = 2$.
4. ✔️ Conclusion: Since $f(x) = x^2$ is continuous on $(-1, 2)$, and the one-sided limits at the endpoints exist and equal the function values, $f(x)$ is continuous on $[-1, 2]$.

Example 2: $f(x) = \sqrt{x}$ on $[0, 4]$

Prove that $f(x) = \sqrt{x}$ is continuous on the closed interval $[0, 4]$.

1. ✅ Continuity on the Open Interval $(0, 4)$: $f(x) = \sqrt{x}$ is continuous for all $x > 0$. Thus, $f(x)$ is continuous for all $x$ in $(0, 4)$.
2. ➡️ Right-Hand Limit at $x = 0$: We need to show that $\lim_{x \to 0^+} f(x) = f(0)$. $$\lim_{x \to 0^+} \sqrt{x} = \sqrt{0} = 0$$ Since $f(0) = \sqrt{0} = 0$, the right-hand limit equals the function value at $x = 0$.
3. ⬅️ Left-Hand Limit at $x = 4$: We need to show that $\lim_{x \to 4^-} f(x) = f(4)$. $$\lim_{x \to 4^-} \sqrt{x} = \sqrt{4} = 2$$ Since $f(4) = \sqrt{4} = 2$, the left-hand limit equals the function value at $x = 4$.
4. ✔️ Conclusion: Since $f(x) = \sqrt{x}$ is continuous on $(0, 4)$, and the one-sided limits at the endpoints exist and equal the function values, $f(x)$ is continuous on $[0, 4]$.

Example 3: $f(x) = \frac{1}{x}$ on $[2, 5]$

Prove that $f(x) = \frac{1}{x}$ is continuous on the closed interval $[2, 5]$.

1. ✅ Continuity on the Open Interval $(2, 5)$: $f(x) = \frac{1}{x}$ is continuous for all $x \neq 0$. Since the interval $(2, 5)$ does not contain 0, $f(x)$ is continuous for all $x$ in $(2, 5)$.
2. ➡️ Right-Hand Limit at $x = 2$: We need to show that $\lim_{x \to 2^+} f(x) = f(2)$. $$\lim_{x \to 2^+} \frac{1}{x} = \frac{1}{2}$$ Since $f(2) = \frac{1}{2}$, the right-hand limit equals the function value at $x = 2$.
3. ⬅️ Left-Hand Limit at $x = 5$: We need to show that $\lim_{x \to 5^-} f(x) = f(5)$. $$\lim_{x \to 5^-} \frac{1}{x} = \frac{1}{5}$$ Since $f(5) = \frac{1}{5}$, the left-hand limit equals the function value at $x = 5$.
4. ✔️ Conclusion: Since $f(x) = \frac{1}{x}$ is continuous on $(2, 5)$, and the one-sided limits at the endpoints exist and equal the function values, $f(x)$ is continuous on $[2, 5]$.

💡 Real-World Applications

• ⚙️ Engineering: Continuity is crucial in modeling physical systems. For example, the stress on a material under load should be a continuous function to avoid sudden failures.
• 🌡️ Physics: Many physical phenomena, such as temperature distribution in a solid, are modeled using continuous functions.
• 📈 Economics: Demand and supply curves are often assumed to be continuous to simplify economic models.

📝 Conclusion

Demonstrating continuity on an interval formally involves verifying continuity at each point within the interval and ensuring that the one-sided limits at the endpoints match the function values. Understanding these principles and practicing with examples will solidify your grasp of this important concept.