1. The problem is to understand the concept of limits and continuity in calculus. 2. 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. A function $f(x)$ is continuous at $x = a$ if three conditions are met: - $f(a)$ is defined. - The limit of $f(x)$ as $x$ approaches $a$ exists. - The limit of $f(x)$ as $x$ approaches $a$ equals $f(a)$. 4. Important rules for limits include: - Limit of a sum is the sum of the limits: $$\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$$ - Limit of a product is the product of the limits: $$\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$$ - Limit of a quotient is the quotient of the limits (if denominator limit is not zero): $$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$$ 5. To find a limit, substitute $x = a$ into $f(x)$ if possible. If substitution leads to an indeterminate form like $\frac{0}{0}$, use algebraic simplification or special techniques like factoring or rationalizing. 6. Example: Find $$\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$$ - Factor numerator: $$x^2 - 4 = (x - 2)(x + 2)$$ - Simplify: $$\frac{(x - 2)(x + 2)}{x - 2} = x + 2$$ for $x \neq 2$ - Substitute $x = 2$: $$2 + 2 = 4$$ - So, $$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$$ 7. This shows the limit exists even though $f(x)$ is not defined at $x=2$ (division by zero). 8. Continuity means no breaks, jumps, or holes in the graph of $f(x)$ at $x=a$. Final answer: Understanding limits and continuity involves evaluating limits using substitution and algebraic techniques and checking the three conditions for continuity at a point.