Fall 2025 Chapter 2-3 Notes: Mastering Limits

Welcome to the Fall 2025 semester! In chapters 2 and 3, we're diving into the fascinating world of limits. These notes will guide you through the essential concepts and techniques you'll need to succeed. Let's embark on this mathematical journey together!

Topics Covered:

  • Using the limit laws to evaluate limits involving algebraic expressions: The limit laws are your best friends when dealing with algebraic expressions. They allow you to break down complex limits into simpler ones. For instance, $$\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$$ and $$\lim_{x \to a} [cf(x)] = c \lim_{x \to a} f(x)$$. These laws make calculations manageable!
  • Finding a limit at a point of continuity: If a function $f(x)$ is continuous at $x=a$, then finding the limit is as easy as direct substitution: $$\lim_{x \to a} f(x) = f(a)$$. Continuity simplifies everything!
  • Finding a limit of a rational function at a removable discontinuity: Removable discontinuities often appear in rational functions where both the numerator and denominator approach zero. Factoring and simplifying can 'remove' the discontinuity, allowing you to evaluate the limit.
  • Finding a limit of a ratio of algebraic functions at a removable discontinuity: Similar to rational functions, identify common factors in the numerator and denominator. Cancel these factors to eliminate the discontinuity and then evaluate the limit.
  • Finding limits for a piecewise-defined function: For piecewise functions, it's crucial to examine the left-hand limit and the right-hand limit at the point where the function's definition changes. If both limits exist and are equal, then the limit exists at that point.
  • Squeeze Theorem: The Squeeze Theorem (also known as the Sandwich Theorem) is a powerful tool. If $f(x) \le g(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$), and if $$\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$$, then $$\lim_{x \to a} g(x) = L$$. This theorem is especially useful for functions involving oscillations. For example, consider $$\lim_{x \to 0} x^2 \sin(\frac{1}{x})$$. Since $-1 \le \sin(\frac{1}{x}) \le 1$, we have $-x^2 \le x^2 \sin(\frac{1}{x}) \le x^2$. As $$\lim_{x \to 0} -x^2 = 0$$ and $$\lim_{x \to 0} x^2 = 0$$, by the Squeeze Theorem, $$\lim_{x \to 0} x^2 \sin(\frac{1}{x}) = 0$$.

Example Problems:

Let's illustrate some of these concepts with examples:

Example 1: Limit Laws

Suppose $\lim_{x \to 2} f(x) = -5$ and $\lim_{x \to 2} g(x) = 8$. Find $\lim_{x \to 2} [f(x) + g(x)]$ and $\lim_{x \to 2} [f(x)g(x)]$.

Solution: Using the limit laws, we have:

  • $$\lim_{x \to 2} [f(x) + g(x)] = \lim_{x \to 2} f(x) + \lim_{x \to 2} g(x) = -5 + 8 = 3$$
  • $$\lim_{x \to 2} [f(x)g(x)] = \lim_{x \to 2} f(x) \cdot \lim_{x \to 2} g(x) = -5 \cdot 8 = -40$$

Example 2: Removable Discontinuity

Find $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$.

Solution: Notice that if we directly substitute $x=1$, we get an indeterminate form $\frac{0}{0}$. We can factor the numerator:

$$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1} (x + 1) = 1 + 1 = 2$$

Example 3: Piecewise Function

Let $$h(x) = \begin{cases} x + 7 & \text{if } x \le -3 \\ \frac{2x - 3}{2x + 3} & \text{if } x > -3 \end{cases}$$. Find $\lim_{x \to -3^-} h(x)$ and $\lim_{x \to -3^+} h(x)$.

Solution:

  • $$\lim_{x \to -3^-} h(x) = \lim_{x \to -3^-} (x + 7) = -3 + 7 = 4$$
  • $$\lim_{x \to -3^+} h(x) = \lim_{x \to -3^+} \frac{2x - 3}{2x + 3} = \frac{2(-3) - 3}{2(-3) + 3} = \frac{-9}{-3} = 3$$

Since the left-hand limit and right-hand limit are not equal, the limit $\lim_{x \to -3} h(x)$ does not exist (DNE).

Keep practicing, and you'll become a limit master! Remember, understanding these concepts is crucial for calculus and beyond. Good luck with your studies!