1. **Stating the problem:** We want to understand the concepts of limit and continuity in calculus. 2. **Definition of Limit:** The limit of a function $f(x)$ as $x$ approaches a value $a$ is the value that $f(x)$ gets closer to as $x$ gets closer to $a$. It is written as $$\lim_{x \to a} f(x) = L$$ where $L$ is the limit. 3. **Important rules for limits:** - Limits can be evaluated by direct substitution if the function is continuous at $a$. - If direct substitution leads to an indeterminate form like $\frac{0}{0}$, we use algebraic simplification, factoring, or special limit laws. 4. **Definition of Continuity:** A function $f(x)$ is continuous at $x = a$ if three conditions are met: - $f(a)$ is defined. - $\lim_{x \to a} f(x)$ exists. - $\lim_{x \to a} f(x) = f(a)$. 5. **Example:** Suppose we want to find $$\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$$ 6. **Step 1:** Direct substitution gives $$\frac{2^2 - 4}{2 - 2} = \frac{4 - 4}{0} = \frac{0}{0}$$ which is indeterminate. 7. **Step 2:** Factor numerator: $$\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}$$ 8. **Step 3:** Cancel common factor: $$\frac{\cancel{(x - 2)}(x + 2)}{\cancel{(x - 2)}} = x + 2$$ 9. **Step 4:** Now substitute $x = 2$: $$2 + 2 = 4$$ 10. **Conclusion:** The limit is 4. Since the simplified function $x + 2$ is continuous at $x=2$, the original function's limit exists and equals 4. This illustrates how limits help us understand function behavior near points, and continuity means no breaks or jumps at that point.