Welcome back to class! In this session, we are moving beyond estimating limits with graphs and tables. We are diving into Section 2.3: Calculating Limits Using the Limit Laws. This is where we start using algebra to solve limit problems precisely.

Note: The video attached to this post is from a previous semester's recording, but it covers the exact same material as our class notes attached below.

1. The Fundamental Limit Laws

Before we can tackle complex functions, we need to understand the rules of the road. The Limit Laws tell us that the limit of a combination of functions is simply the combination of their limits (provided those limits exist). For example, the limit of a sum is the sum of the limits:

$$\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$$

This leads us to the Direct Substitution Property. If $f$ is a polynomial or a rational function and $a$ is in the domain of $f$, then:

$$\lim_{x \to a} f(x) = f(a)$$

Simply put: try plugging the number in first! If you get a valid number (like in our example $\lim_{x \to 5} (2x^2 - 3x + 4) = 39$), you are done.

2. Handling Indeterminate Forms ($0/0$)

The real calculus work begins when direct substitution gives you $\frac{0}{0}$. This is called an indeterminate form. It doesn't mean the limit doesn't exist; it means we have more work to do to find it. We usually employ two main algebraic strategies:

  • Factoring: If you have a polynomial over a polynomial, try to factor and cancel common terms.
    Example from notes: $$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2$$
  • Conjugates: If you see a square root involved in a fraction that results in $0/0$, multiply the numerator and the denominator by the conjugate.
    Example from notes: For $\lim_{t \to 0} \frac{\sqrt{t^2+9}-3}{t^2}$, we multiply by $\frac{\sqrt{t^2+9}+3}{\sqrt{t^2+9}+3}$. This clears the root from the numerator and allows us to simplify the expression to find the limit is $\frac{1}{6}$.

3. Piecewise Functions and One-Sided Limits

Remember that for a general limit $\lim_{x \to a} f(x)$ to exist, the left-hand limit and the right-hand limit must be equal. This is crucial when dealing with piecewise functions.

$$\lim_{x \to a} f(x) = L \quad \text{if and only if} \quad \lim_{x \to a^-} f(x) = L = \lim_{x \to a^+} f(x)$$

If the left side approaches a different value than the right side (a "jump" in the graph), the limit Does Not Exist (DNE).

4. The Squeeze Theorem

Sometimes we encounter wild functions that oscillate rapidly, like $f(x) = x^2 \sin(\frac{1}{x})$. We can't use standard laws here. Instead, we use the Squeeze Theorem.

If we can find two functions, one above ($h(x)$) and one below ($g(x)$), such that our difficult function is "squeezed" between them, and the limits of the top and bottom functions match at $a$, then the limit of our middle function must be the same.

Since $-1 \le \sin(\frac{1}{x}) \le 1$, we can multiply by $x^2$ to get:

$$-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$$

Because $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} (x^2) = 0$, the function in the middle is forced to be $0$.

Make sure to download the PDF notes below to see the step-by-step solutions for the conjugate and expanding examples. Keep practicing those algebra skills—they are the key to unlocking Calculus!