Welcome back to Calculus 1! In our September 6th class, we tackled Section 2-3: Calculating Limits Using the Limit Laws. While tables and graphs (from Section 2-2) give us a good estimate, this section is all about precision. We learned how to use algebraic properties to find the exact value of a limit without having to plot points every time.

1. The Limit Laws

We started by defining the fundamental rules that allow us to break complex limits into simpler parts. If the limits for $f(x)$ and $g(x)$ exist as $x \to a$, we can manipulate them using arithmetic operations. Key laws include:

  • Sum/Difference Law: $\lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)$
  • Constant Multiple Law: $\lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x)$
  • Quotient Law: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}$ (provided the denominator is not zero!)

2. The Direct Substitution Property

For many continuous functions—specifically polynomials and rational functions within their domains—finding a limit is straightforward. You simply plug in the value.

For example, if we have a polynomial function, $\lim_{x \to a} f(x) = f(a)$. This is always our first step when evaluating a limit.

3. When Direct Substitution Fails (0/0)

The most interesting problems occur when plugging in $x=a$ results in $\frac{0}{0}$. This is called an indeterminate form. It doesn't mean the limit doesn't exist; it means we have more work to do! We covered several algebraic techniques to resolve this:

  • Factoring: As seen in the class notes (Example 3 and 16), if we have a rational function like $\frac{x^2-1}{x-1}$, we can factor the numerator into $(x-1)(x+1)$ and cancel the common term.
  • Expanding: Sometimes you need to do the algebra first. In Example 5, we looked at $\lim_{h \to 0} \frac{(3+h)^2 - 9}{h}$. By expanding the quadratic, the terms cancel out nicely, allowing us to evaluate the limit.
  • Conjugates: When dealing with square roots (like in Example 6), we multiply the numerator and denominator by the conjugate to eliminate the root and simplify the expression.

4. The Squeeze Theorem

Finally, we looked at the Squeeze Theorem (sometimes called the Sandwich Theorem). This is used when a function is trapped between two other functions with the same limit.

If $f(x) \le g(x) \le h(x)$ near $a$, and: $$\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$$ Then, it must be true that $\lim_{x \to a} g(x) = L$. We demonstrated this with the classic example of $\lim_{x \to 0} x^2 \sin(1/x) = 0$.

Be sure to review the attached PDF for the step-by-step solutions to the practice problems we did in class, especially the "Difference of Squares" factoring examples!