Welcome back. In our last lecture, we developed an intuition for what a limit is by using tables and graphs to estimate its value. While this is a great way to understand the concept, it can be slow and sometimes imprecise. Today, we move from estimation to calculation. We will learn a set of powerful algebraic rules called the Limit Laws that allow us to calculate limits exactly and efficiently. 🚀
Suppose that $c$ is a constant and that the limits $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ both exist. Then we have the following laws:
These laws are built upon two foundational limits: $\lim_{x \to a} c = c$ and $\lim_{x \to a} x = a$.
Use the Limit Laws to evaluate $\lim_{x \to 3} (2x^2 - 5x + 1)$.
$$ \begin{align*} \lim_{x \to 3} (2x^2 - 5x + 1) &= \lim_{x \to 3} (2x^2) - \lim_{x \to 3} (5x) + \lim_{x \to 3} (1) & & \text{ (Laws 1 & 2)} \\ &= 2 \lim_{x \to 3} (x^2) - 5 \lim_{x \to 3} (x) + \lim_{x \to 3} (1) & & \text{ (Law 3)} \\ &= 2 \left(\lim_{x \to 3} x\right)^2 - 5 \lim_{x \to 3} (x) + \lim_{x \to 3} (1) & & \text{ (Law 6)} \\ &= 2(3)^2 - 5(3) + 1 & & \text{ (Direct Substitution)} \\ &= 18 - 15 + 1 = 4 \end{align*} $$
This detailed process confirms that for any polynomial, we can find the limit by simply substituting the value of $x$.
Problem: Given that $\lim_{x \to a} f(x) = 9$ and $\lim_{x \to a} g(x) = -2$, find the value of $\lim_{x \to a} \frac{2f(x) - g(x)}{\sqrt{f(x)}}$.
So far, we've been looking at the overall limit of a function as $x$ approaches a value $a$. But what if we only care about the path from one side? The concept of a one-sided limit allows us to be more precise about a function's behavior.
We use the notation $x \to a^-$ to mean "$x$ approaches $a$ from the left" (through values less than $a$) and $x \to a^+$ to mean "$x$ approaches $a$ from the right" (through values greater than $a$).
The two-sided limit $\lim_{x \to a} f(x) = L$ exists if and only if both one-sided limits exist and are equal:
$$ \lim_{x \to a^-} f(x) = L \quad \text{and} \quad \lim_{x \to a^+} f(x) = L $$
In simple terms: The path from the left and the path from the right must meet at the exact same point for the overall limit to exist.
Consider the graph of the piecewise function $g(x)$ shown below.
Let's find the limits as $x \to 2$.
Let's find the limit of $f(x) = \frac{x-3}{|x-3|}$ as $x \to 3$.
(a) Limit from the right ($x \to 3^+$):
If $x > 3$, then $x-3$ is positive, so $|x-3| = x-3$. Therefore:
$$ \lim_{x \to 3^+} \frac{x-3}{|x-3|} = \lim_{x \to 3^+} \frac{x-3}{x-3} = \lim_{x \to 3^+} 1 = 1 $$
(b) Limit from the left ($x \to 3^-$):
If $x < 3$, then $x-3$ is negative, so $|x-3| = -(x-3)$. Therefore:
$$ \lim_{x \to 3^-} \frac{x-3}{|x-3|} = \lim_{x \to 3^-} \frac{x-3}{-(x-3)} = \lim_{x \to 3^-} -1 = -1 $$
Since the left-hand limit ($-1$) does not equal the right-hand limit ($1$), the overall limit does not exist.
Problem: Let $f(x) = \frac{x+5}{x-2}$. Find the one-sided limits as $x$ approaches 2.
For polynomials and rational functions, calculating a limit is as simple as plugging in the value, as long as it doesn't make the denominator zero. This is called the Direct Substitution Property.
However, if direct substitution gives you the indeterminate form $\frac{0}{0}$, you must perform more steps. The goal is always to algebraically manipulate the function so that you can cancel the term causing the zero in the denominator.
Calculate the limit $\lim_{x \to -2} \frac{x^2+5x+6}{x+2}$.
Direct substitution gives $\frac{0}{0}$. We must simplify by factoring the numerator:
$$ \lim_{x \to -2} \frac{(x+2)(x+3)}{x+2} = \lim_{x \to -2} (x+3) = -2 + 3 = 1 $$
Problem: Let $f(x) = x^2$.
What if we can't calculate a limit directly, but we can trap or "squeeze" our difficult function between two simpler functions whose limits are the same? This is the idea behind the Squeeze Theorem.
If $f(x) \le g(x) \le h(x)$ when $x$ is near $a$, and $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} g(x) = L$.
Show that $\lim_{x \to 0} x^2 \sin(\frac{1}{x}) = 0$.
We cannot use the Product Law because $\lim_{x \to 0} \sin(\frac{1}{x})$ does not exist. But we know that $-1 \le \sin(\frac{1}{x}) \le 1$. Multiplying by $x^2$ (which is non-negative) gives:
$$ -x^2 \le x^2 \sin(\frac{1}{x}) \le x^2 $$
Since $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} (x^2) = 0$, our function is squeezed between two functions that both approach 0. By the Squeeze Theorem, our limit must also be 0.
For the following problems, find the limit using algebraic techniques. It's important to know how to do this for an exam. For these problems, you can also use Desmos to plot the graph and visually confirm your answer.
Given $\lim_{x \to 2} f(x) = 5$ and $\lim_{x \to 2} g(x) = -3$, find $\lim_{x \to 2} [4f(x) - g(x)]$.
Find the limit: $\lim_{u \to -2} \sqrt{9-u^3 + 2u^2}$.
Find the limit: $\lim_{x \to 3} \frac{1/x - 1/3}{x-3}$.
Today we established a powerful set of rules for calculating limits algebraically. The Limit Laws, combined with techniques like factoring and conjugates, allow us to solve for exact limits without relying on tables or graphs. We also introduced the Squeeze Theorem, a clever tool for dealing with tricky functions. These methods form the bedrock for our next major topic: continuity.