Welcome back, everyone. In our last lecture, we saw that the problems of finding a tangent slope and an instantaneous velocity both relied on the same core strategy: approximating a value by looking at what happens as we get closer and closer to a specific point. Today, we give this powerful idea its formal name: the limit.
Let's start with an intuitive definition. We say that the limit of a function $f(x)$ as $x$ approaches a number $a$ is equal to a number $L$ if we can make the values of $f(x)$ as close to $L$ as we like by taking $x$ to be sufficiently close to $a$ (on either side of $a$) but not equal to $a$.
We write this using the following notation:
The most important part of this definition is that we do not care about the value of the function at $x=a$. The function might be defined at $a$, or it might not. The limit is only concerned with the behavior of the function "near" $a$.
A common scenario is trying to find a limit by "plugging in" the value of $a$. Sometimes this works, but often it gives us the form $\frac{0}{0}$. This is called an indeterminate form. It doesn't mean the limit is 0 or that it doesn't exist. It's a signal that we must do more work, either by investigating numerically with a table of values or by using algebra to simplify the function.
Use a table of values to estimate the value of $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.
First, notice that if we try to substitute $x=2$ into the function, we get $\frac{2^2 - 4}{2 - 2} = \frac{0}{0}$. This is our indeterminate form, so we must investigate further. Let's create a table of values for $x$ approaching 2 from both sides.
| $x$ (from the left) | $f(x) = \frac{x^2 - 4}{x - 2}$ | $x$ (from the right) | $f(x) = \frac{x^2 - 4}{x - 2}$ |
|---|---|---|---|
| 1.9 | 3.9 | 2.1 | 4.1 |
| 1.99 | 3.99 | 2.01 | 4.01 |
| 1.999 | 3.999 | 2.001 | 4.001 |
The table shows that as $x$ gets closer to 2, $f(x)$ gets closer to 4. We can also see this by simplifying the function algebraically for $x \neq 2$: $f(x) = \frac{(x-2)(x+2)}{x-2} = x+2$. The graph of our function is the line $y=x+2$ with a hole at $x=2$. The limit is the height of that hole. Therefore, we conclude:
Find the value of $\lim_{x \to 9} \frac{\sqrt{x} - 3}{x-9}$.
Direct substitution of $x=9$ gives $\frac{0}{0}$. This is an indeterminate form. Here we can use algebra. The key is to multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x}+3$.
$$ \lim_{x \to 9} \left( \frac{\sqrt{x} - 3}{x-9} \right) \cdot \left( \frac{\sqrt{x} + 3}{\sqrt{x} + 3} \right) = \lim_{x \to 9} \frac{x-9}{(x-9)(\sqrt{x}+3)} $$
Since $x$ is only approaching 9, $x \neq 9$, so we can cancel the $(x-9)$ terms.
$$ \lim_{x \to 9} \frac{1}{\sqrt{x}+3} = \frac{1}{\sqrt{9}+3} = \frac{1}{6} $$
Problem: Use a table of values to estimate $\lim_{x \to 0} \frac{e^{2x} - 1}{x}$.
Sometimes a function behaves differently depending on whether we approach a point from the left or the right. This requires a more precise tool.
We write $\lim_{x \to a^-} f(x) = L$ for the left-hand limit (approaching from values less than $a$).
We write $\lim_{x \to a^+} f(x) = L$ for the right-hand limit (approaching from values greater than $a$).
This leads to a crucial condition for the existence of a standard, two-sided limit:
$$ \lim_{x \to a} f(x) = L \quad \text{if and only if} \quad \lim_{x \to a^-} f(x) = L \quad \text{and} \quad \lim_{x \to a^+} f(x) = L $$
In other words, for a limit to exist, the function must be approaching the same value from both the left and the right.
Consider the piecewise function $f(x) = \begin{cases} x+2 & \text{if } x < 1 \\ x^2 - 3 & \text{if } x \geq 1 \end{cases}$. Find the left-hand limit, the right-hand limit, and the overall limit as $x \to 1$.
Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+2) = 1+2 = 3$.
Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2-3) = 1^2 - 3 = -2$.
Since the left-hand limit (3) is not equal to the right-hand limit (-2), the overall limit $\lim_{x \to 1} f(x)$ does not exist. This is called a jump discontinuity.
Find the one-sided limits of $f(x) = \frac{|x-2|}{x-2}$ as $x \to 2$.
We can write the absolute value as a piecewise function: $|x-2| = \begin{cases} x-2 & \text{if } x \geq 2 \\ -(x-2) & \text{if } x < 2 \end{cases}$.
Left-hand limit ($x<2$): $\lim_{x \to 2^-} \frac{-(x-2)}{x-2} = \lim_{x \to 2^-} (-1) = -1$.
Right-hand limit ($x>2$): $\lim_{x \to 2^+} \frac{x-2}{x-2} = \lim_{x \to 2^+} (1) = 1$.
Since the one-sided limits differ, the overall limit does not exist.
Problem: Let $f(x) = \begin{cases} \sqrt{9-x^2} & \text{if } -3 \le x < 3 \\ x-3 & \text{if } x \ge 3 \end{cases}$. Find $\lim_{x \to 3^-} f(x)$, $\lim_{x \to 3^+} f(x)$, and $\lim_{x \to 3} f(x)$.
What happens if the values of $f(x)$ grow without bound as $x$ approaches $a$? We call this an infinite limit. This often happens when the denominator of a fraction approaches zero while the numerator approaches a non-zero number.
We write $\lim_{x \to a} f(x) = \infty$ if the values of $f(x)$ grow arbitrarily large and positive. When this occurs, the line $x=a$ is a vertical asymptote.
When we say a limit equals $\infty$, we are describing "why" the limit does not exist. Remember, $\infty$ is a concept, not a real number.
Find the limit of $f(x) = \frac{1}{(x-3)^2}$ as $x \to 3$.
As $x$ approaches 3 (from either side), the term $(x-3)^2$ becomes a very small positive number. Dividing 1 by a very small positive number results in a very large positive number. Therefore, $\lim_{x \to 3} \frac{1}{(x-3)^2} = \infty$. The line $x=3$ is a vertical asymptote.
Find the limit of $f(x) = \frac{x}{(x+2)^2}$ as $x \to -2$.
As $x \to -2$, the numerator $x$ approaches -2. The denominator $(x+2)^2$ approaches 0 and will always be positive. Dividing a negative number by a very small positive number results in a large negative number. Therefore, $\lim_{x \to -2} \frac{x}{(x+2)^2} = -\infty$. The line $x=-2$ is a vertical asymptote.
Problem: Find the one-sided limits of $f(x) = \frac{x+1}{x-4}$ as $x \to 4$.
To summarize, a limit $\lim_{x \to a} f(x)$ fails to exist for three main reasons:
Find the limit using algebraic techniques, and check your answer by plotting the graph in Desmos.
$$ \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} $$
Find the limit using algebraic techniques, and check your answer by plotting the graph in Desmos.
$$ \lim_{x \to 3^-} \frac{x^2 + 4x}{x^2 - 2x - 3} $$
Find the limit using algebraic techniques, and check your answer by plotting the graph in Desmos.
$$ \lim_{x \to -5} \frac{x^2+3x-10}{x+5} $$
Today we have built a solid intuition for what a limit is, how to investigate it numerically and graphically, and the different ways a limit can exist or fail to exist. In our next lecture, we will move beyond this estimation-based approach and learn the algebraic rules—the Limit Laws—that will allow us to calculate limits precisely and efficiently.