Welcome back to Professor Baker's Math Class! In our September 20th session, we conducted a thorough review for the upcoming Chapter 1 & 2 Test. This session is crucial as it bridges the gap between understanding function definitions and mastering the concept of limits.

Below you will find the blank review sheet to practice on your own, along with the detailed class notes showing the step-by-step solutions.

Download the Class Materials:

Key Concepts Covered

Here is a breakdown of the essential topics we discussed in class, which will be vital for the test:

1. The Rule of Four

We started by looking at functions through four distinct lenses: Verbal, Numerical, Graphical, and Algebraic. In Problem 1, we examined the function $f(x) = x^2 + 2x - 4$.

Class Note: We discussed how this specific parabola could model the path of a diver. Being able to translate between a table of values, a graph, and an equation is a fundamental skill for calculus.

2. Graphical Limits & Continuity

We practiced evaluating limits simply by looking at a graph. Remember the Golden Rule of limits:

$$ \lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x) \implies \lim_{x \to c} f(x) \text{ Does Not Exist (DNE)} $$

We also reviewed how to sketch graphs based on continuity descriptions, specifically distinguishing between a Removable Discontinuity (a hole) and a Jump Discontinuity (a break in the graph).

3. Algebraic Limits & Common Pitfalls

When direct substitution results in an indeterminate form like $\frac{0}{0}$, we must use algebra to find the solution. We practiced two main techniques:

  • Factoring: In Problem 6, we solved $\lim_{t \to 4} \frac{t^2-2t-8}{t-4}$ by factoring the numerator to cancel out the $(t-4)$ term, revealing the limit is 6.
  • Expanding: In Problem 7, we tackled $\lim_{h \to 0} \frac{(h-3)^2-9}{h}$.

⚠️ Algebra Warning: As noted in the slides, be careful when expanding binomials! A common mistake is to think $(h-3)^2 = h^2 - 9$. This is incorrect.
The correct expansion is $(h-3)(h-3) = h^2 - 6h + 9$.

4. Limits at Infinity

Finally, we looked at the end behavior of functions. For rational functions like the one in Problem 10:

$$ \lim_{t \to -\infty} \frac{3t^2+t}{t^3-4t+1} $$

Since the degree of the denominator ($t^3$) is greater than the degree of the numerator ($t^2$), the function settles toward 0 as $t$ gets incredibly large (or small).

Review the attached notes carefully, pay attention to your algebra, and good luck studying! You've got this!