Chapter 3: Intuition for Limits
The idea of “getting arbitrarily close”
Beginner
3.1 What Is a Limit?
In the previous chapter, we used the expression “$h \to 0$” to find the slope of a tangent line. This is connected to the idea of a limit.
$\displaystyle\lim_{x \to a} f(x) = L$ means:
“As $x$ gets arbitrarily close to $a$, $f(x)$ gets arbitrarily close to $L$.”
The key point is that a limit describes how $f(x)$ behaves when $x$ is close to $a$ (but $x\neq a$), rather than its value when $x=a$. So whether $f(a)$ is defined does not affect whether the limit exists.
3.2 Limits of Sequences
To understand limits, let us first look at the limit of a sequence.
Example: The Limit of 1/n
Consider the sequence $a_n = \dfrac{1}{n}$.
| n | 1 | 2 | 10 | 100 | 1000 | ... |
|---|---|---|---|---|---|---|
| $a_n = 1/n$ | 1 | 0.5 | 0.1 | 0.01 | 0.001 | ... |
As $n$ gets larger, $a_n$ gets arbitrarily close to 0.
$$\lim_{n \to \infty} \dfrac{1}{n} = 0$$$\dfrac{1}{n}$ gets close to 0, but never reaches 0.
No matter how large $n$ is, $\dfrac{1}{n} > 0$.
3.3 Limits of Functions
Now that we have looked at limits of sequences, let us consider limits of functions.
Example 1: A Simple Limit
$$\lim_{x \to 2} (x + 3)$$What happens to $x + 3$ as $x$ approaches 2?
| x | 1.9 | 1.99 | 1.999 | → | 2.001 | 2.01 | 2.1 |
|---|---|---|---|---|---|---|---|
| x + 3 | 4.9 | 4.99 | 4.999 | → 5 | 5.001 | 5.01 | 5.1 |
Clearly, $x + 3$ approaches 5.
$$\lim_{x \to 2} (x + 3) = 5$$Example 2: A Slightly More Involved Limit
$$\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$$There is a problem here. Substituting $x = 2$ gives:
$$\dfrac{2^2 - 4}{2 - 2} = \dfrac{0}{0}$$$\dfrac{0}{0}$ is called an indeterminate form. This form alone does not tell us the value of the limit.
But a limit is about the value as $x$ approaches 2, not the value at $x = 2$.
Step 1: Factor the numerator
$$x^2 - 4 = (x+2)(x-2)$$Step 2: Cancel the common factor (when $x \neq 2$)
$$\dfrac{x^2 - 4}{x - 2} = \dfrac{(x+2)(x-2)}{x-2} = x + 2 \quad (x \neq 2)$$Step 3: Find the limit
$$\lim_{x \to 2} (x + 2) = 2 + 2 = 4$$The graph shows that the function is undefined at $x = 2$, but as $x$ approaches 2, the function values approach 4.
3.4 Limit Laws
If the limits of the functions $f(x)$ and $g(x)$ both exist as finite values, the following laws apply.
Suppose $\displaystyle\lim_{x \to a} f(x) = L$ and $\displaystyle\lim_{x \to a} g(x) = M$ (where $L,M$ are finite real numbers). Then:
1. Sum Law
$$\lim_{x \to a} [f(x) + g(x)] = L + M$$2. Difference Law
$$\lim_{x \to a} [f(x) - g(x)] = L - M$$3. Product Law
$$\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$$4. Quotient Law (when $M \neq 0$)
$$\lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{L}{M}$$5. Constant Multiple Law
$$\lim_{x \to a} [c \cdot f(x)] = c \cdot L$$3.5 The Limit as $h \to 0$
The limits used in derivatives often have the form $h \to 0$.
Example: A Limit Used to Find a Derivatives
When we found the derivative of $f(x) = x^2$, we calculated this limit:
$$\lim_{h \to 0} \dfrac{(x+h)^2 - x^2}{h}$$Step 1: Expand $(x+h)^2$
$$(x+h)^2 = x^2 + 2xh + h^2$$Step 2: Simplify the numerator
$$(x+h)^2 - x^2 = x^2 + 2xh + h^2 - x^2 = 2xh + h^2$$Step 3: Divide by $h$
$$\dfrac{(x+h)^2 - x^2}{h} = \dfrac{2xh + h^2}{h}$$Step 4: Factor $h$ from the numerator
$$\dfrac{h(2x + h)}{h} = 2x + h \quad (h \neq 0)$$Step 5: Find the limit as $h \to 0$
$$\lim_{h \to 0} (2x + h) = 2x + 0 = 2x$$Therefore, $(x^2)' = 2x$.
3.6 When a Limit Does Not Exist
Not every limit exists.
Example 1: Divergence (Growing Without Bound)
$$\lim_{x \to 0} \dfrac{1}{x^2}$$As $x$ approaches 0, $\dfrac{1}{x^2}$ grows without bound.
When a function’s values grow without bound like this, we say that they “diverge to $+\infty$.” We may also write $\displaystyle\lim_{x \to 0} \dfrac{1}{x^2} = +\infty$. However, this does not mean that a limit exists in the usual sense of converging to a finite real number.
Example 2: Oscillation (No Single Value)
$$\lim_{x \to 0} \sin\left(\dfrac{1}{x}\right)$$As $x$ approaches 0, the absolute value of $\dfrac{1}{x}$ grows without bound, and the value of $\sin$ keeps oscillating between $-1$ and $1$. So it does not approach a single value.
- The function does not approach a finite value and diverges to $+\infty$ or $-\infty$
- The function oscillates and does not approach a single value
- The left-hand and right-hand limits are different
3.7 One-Sided Limits
In the limits we have considered so far, $x$ approached $a$ from both sides. We can also consider limits where $x$ approaches from just one side.
Right-hand limit (approaching from the right):
$$\lim_{x \to a^+} f(x) \quad \text{or} \quad \lim_{x \to a+0} f(x)$$The limit as $x$ approaches $a$ from the side greater than $a$.
Left-hand limit (approaching from the left):
$$\lim_{x \to a^-} f(x) \quad \text{or} \quad \lim_{x \to a-0} f(x)$$The limit as $x$ approaches $a$ from the side less than $a$.
The necessary and sufficient condition for $\displaystyle\lim_{x \to a} f(x)$ to exist as a finite real number is that the left-hand and right-hand limits both exist as finite real numbers and are equal:
$$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$$In other words, the left-hand and right-hand limits must be equal.
Example: When the Left-Hand and Right-Hand Limits Differ
Consider the sign function $\text{sgn}(x) = \begin{cases} 1 & (x \gt 0) \\ 0 & (x = 0) \\ -1 & (x \lt 0) \end{cases}$.
In this case:
$$\lim_{x \to 0^-} \text{sgn}(x) = -1, \quad \lim_{x \to 0^+} \text{sgn}(x) = 1$$Since the left-hand and right-hand limits are different, $\displaystyle\lim_{x \to 0} \text{sgn}(x)$ does not exist.
- Checking differentiability (for example, $|x|$, which we will study in Chapter 4)
- Classifying points of discontinuity
- Functions with restricted domains, such as $\ln x$ and $\sqrt{x}$
Chapter Summary
- Limit: The value a function approaches as $x$ gets close to $a$
- The limit is different from the value at $x = a$ (it describes approaching, not reaching)
- Limit laws: Limits of sums, differences, products, and quotients can be calculated as usual
- The indeterminate form $\dfrac{0}{0}$ can be resolved by factoring and other methods
- Derivatives use limits as $h \to 0$
- A function may not converge to a finite value (it may diverge to infinity or oscillate)
- One-sided limits: $\displaystyle\lim_{x \to a^+}$ (right-hand limit), $\displaystyle\lim_{x \to a^-}$ (left-hand limit)
- A limit exists if and only if the left-hand and right-hand limits are equal
Frequently Asked Questions
Is “$x \to a$” the same as “$x = a$”?
No. They are completely different. $\lim_{x \to a} f(x)$ describes what happens as $x$ gets arbitrarily close to $a$, and is unrelated to the value $f(a)$ at $x = a$. A limit can be found even at a point where $f(a)$ is undefined. For example, $f(x) = \dfrac{x^2-1}{x-1}$ is undefined at $x=1$, but $\lim_{x \to 1} f(x) = 2$.
How are limits and continuity related?
A function $f$ is continuous at $x=a$ if $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and $\lim_{x \to a} f(x)=f(a)$. In other words, a function is continuous at a point when its limit there equals its value at that point. The function is not continuous there if the limit exists but does not equal $f(a)$, or if $f(a)$ is undefined.
How can you resolve the indeterminate form $\dfrac{0}{0}$?
Common methods include (1) factoring (for example, $\dfrac{x^2-9}{x-3} = x+3$), (2) rationalizing when radicals are involved, and (3) L’Hôpital’s rule, which is studied in college. In basic problems, look for and cancel a common factor in the numerator and denominator.
Practice Problems
Problem 1 (Basic): Find $\displaystyle\lim_{x \to 2} (3x - 1)$.
Hint
The function is continuous, so you can substitute directly. $3 \cdot 2 - 1 = 5$.
Problem 2 (Standard): Find $\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$.
Hint
Factor $x^2 - 9 = (x-3)(x+3)$. Then $x+3 \to 6$.
Problem 3 (Challenge): Find $\displaystyle\lim_{h \to 0} \dfrac{(2+h)^2 - 4}{h}$. Then explain what this means.
Hint
Expanding gives $\dfrac{4h + h^2}{h} = 4 + h \to 4$. This is the derivative at the point $x=2$ for $f(x) = x^2$ (the slope of the tangent line).