The Idea of a Limit
Understand the fundamental concept that makes calculus possible
You already understand limits better than you think. Every time you watch a car’s speedometer, you are witnessing a limit in action. The speedometer does not tell you how far you traveled divided by how long you have been driving (that would be your average speed for the whole trip). Instead, it tells you how fast you are going right now, at this very instant. But here is the puzzle: an “instant” has no duration. How can you measure speed when no time passes? The answer is limits, and it turns out this question is one that humans struggled with for over two thousand years before calculus finally provided a satisfying answer.
The good news is that the core idea is remarkably intuitive. When we ask “what is the limit?” we are really just asking “what value is this thing getting closer and closer to?” You do this naturally all the time. If you are walking toward a wall, you get closer and closer to it. If someone asked “where are you heading?” you would point at the wall, not at any of the intermediate positions you happen to occupy along the way. That is limit-thinking: identifying the destination, not the journey.
Core Concepts
What is a Limit?
A limit describes what value a function approaches as its input approaches some particular number. The key word here is approaches. We are not asking what the function equals at that point; we are asking what it gets arbitrarily close to as we get closer and closer to that point.
Think of it like this: imagine you are walking toward a doorway. At each step, you cover half the remaining distance. After your first step, you are halfway there. After your second step, you have covered three-quarters of the distance. Then seven-eighths, then fifteen-sixteenths, and so on. You never actually reach the doorway in any finite number of steps, but it is perfectly clear where you are heading. The doorway is the limit of your position.
This is not just a thought experiment. The ancient Greek philosopher Zeno posed several famous paradoxes about motion and infinity, and they haunted mathematicians for centuries. How can Achilles ever catch a tortoise if he must first reach where the tortoise was, by which time the tortoise has moved? The answer is limits: the infinite sum of all those ever-smaller distances has a finite limit.
Mathematically, we write:
$$\lim_{x \to a} f(x) = L$$
This reads as “the limit of $f(x)$ as $x$ approaches $a$ equals $L$.” It means that as $x$ gets closer and closer to $a$ (but not necessarily equaling $a$), the function values $f(x)$ get closer and closer to $L$.
The Crucial Distinction: Near vs. At
Here is something that trips people up at first: limits are about what happens near a point, not at the point. This distinction matters enormously.
Consider the function:
$$f(x) = \frac{x^2 - 1}{x - 1}$$
If you try to plug in $x = 1$, you get $\frac{0}{0}$, which is undefined. The function literally has no value at $x = 1$. But watch what happens as we approach $x = 1$:
| $x$ | $f(x)$ |
|---|---|
| 0.9 | 1.9 |
| 0.99 | 1.99 |
| 0.999 | 1.999 |
| 1.001 | 2.001 |
| 1.01 | 2.01 |
| 1.1 | 2.1 |
The function values are clearly approaching 2, even though the function is undefined at $x = 1$. We write:
$$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2$$
This is why limits are so powerful. They let us talk about the behavior of a function at a point where the function might not even be defined.
(If you are curious why this works, notice that $x^2 - 1 = (x+1)(x-1)$, so for $x \neq 1$, we have $\frac{x^2-1}{x-1} = \frac{(x+1)(x-1)}{x-1} = x + 1$. As $x \to 1$, this approaches $1 + 1 = 2$.)
Finding Limits from Tables
One practical way to estimate a limit is to make a table of values, approaching the target from both sides. This is exactly what we did above. The key is to look for a pattern: are the function values settling down toward some specific number?
When making a table, approach the target from both directions:
- From the left (values smaller than the target)
- From the right (values larger than the target)
If both sides approach the same value, that is your limit.
Finding Limits from Graphs
Reading a limit from a graph is even more intuitive. You trace the curve with your eye, approaching the target $x$-value from both the left and the right. Where is the function heading? That height is the limit.
Important: look at where the curve is heading, not necessarily where there might be a dot or hole on the graph. A function can have a hole at a point (meaning it is undefined there) but still have a perfectly well-defined limit. Conversely, a function can be defined at a point but approach different values from different directions.
One-Sided Limits
Sometimes we want to consider what happens when we approach from just one direction. These are called one-sided limits.
- The left-hand limit $\lim_{x \to a^-} f(x)$ describes what $f(x)$ approaches as $x$ approaches $a$ from values less than $a$ (from the left on a number line).
- The right-hand limit $\lim_{x \to a^+} f(x)$ describes what $f(x)$ approaches as $x$ approaches $a$ from values greater than $a$ (from the right).
The little minus sign in $a^-$ means “from below” or “from the left,” and the little plus sign in $a^+$ means “from above” or “from the right.”
A two-sided limit exists if and only if:
- The left-hand limit exists
- The right-hand limit exists
- They are equal to each other
$$\lim_{x \to a} f(x) = L \quad \text{if and only if} \quad \lim_{x \to a^-} f(x) = L \quad \text{and} \quad \lim_{x \to a^+} f(x) = L$$
When Limits Do Not Exist
Not every function has a limit at every point. Here are the main ways a limit can fail to exist:
1. Jump discontinuity: The left and right limits exist but are different. Picture a step function that jumps from one value to another.
2. Infinite discontinuity: The function shoots off to positive or negative infinity. For example, $\lim_{x \to 0} \frac{1}{x^2} = +\infty$. (Technically, we say this limit “does not exist” as a finite number, though we can describe the behavior as “tending to infinity.”)
3. Oscillating behavior: The function oscillates more and more wildly without settling down. We will see an example of this shortly.
When a limit does not exist, we write “DNE” (does not exist).
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| $\lim_{x \to a} f(x)$ | The value $f(x)$ approaches as $x$ approaches $a$ | $\lim_{x \to 2} (3x) = 6$ |
| Left-hand limit | Limit as $x$ approaches from below (smaller values) | $\lim_{x \to a^-} f(x)$ |
| Right-hand limit | Limit as $x$ approaches from above (larger values) | $\lim_{x \to a^+} f(x)$ |
| Limit exists | Both one-sided limits exist and are equal | $\lim_{x \to 3} x^2 = 9$ |
| DNE | “Does not exist” - the limit is not a single finite number | |
| Approaches | Gets arbitrarily close to (but may not equal) | “$x$ approaches 5” means $x \to 5$ |
| Tends to infinity | Grows without bound | $\lim_{x \to 0^+} \frac{1}{x} = +\infty$ |
Examples
Find $\lim_{x \to 3} (2x + 1)$.
Solution:
Let us build a table of values approaching $x = 3$ from both sides:
| $x$ | $2x + 1$ |
|---|---|
| 2.9 | 6.8 |
| 2.99 | 6.98 |
| 2.999 | 6.998 |
| 3.001 | 7.002 |
| 3.01 | 7.02 |
| 3.1 | 7.2 |
From the left side, the values approach 7. From the right side, the values approach 7.
Both sides agree, so: $$\lim_{x \to 3} (2x + 1) = 7$$
Of course, you could also just substitute $x = 3$ directly: $2(3) + 1 = 7$. For “nice” functions like polynomials, the limit equals the function value. But the table method helps build intuition and works even when direct substitution does not.
Suppose you are given the graph of a function $g(x)$ with the following features:
- As $x$ approaches 2 from the left, the curve heads toward height 5
- As $x$ approaches 2 from the right, the curve heads toward height 5
- At $x = 2$, there is a hole in the graph (the function is undefined there)
What is $\lim_{x \to 2} g(x)$?
Solution:
Even though $g(2)$ is undefined (there is a hole), we can still find the limit by looking at where the curve is heading.
From both directions, the function approaches the height 5.
Therefore: $$\lim_{x \to 2} g(x) = 5$$
This illustrates the key principle: limits describe where the function is going, not necessarily where it is.
Consider the piecewise function: $$f(x) = \begin{cases} x + 1 & \text{if } x < 2 \ x^2 - 1 & \text{if } x \geq 2 \end{cases}$$
Find $\lim_{x \to 2^-} f(x)$, $\lim_{x \to 2^+} f(x)$, and determine if $\lim_{x \to 2} f(x)$ exists.
Solution:
Left-hand limit: As $x$ approaches 2 from the left, we use the formula $f(x) = x + 1$: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 1) = 2 + 1 = 3$$
Right-hand limit: As $x$ approaches 2 from the right, we use the formula $f(x) = x^2 - 1$: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 - 1) = 4 - 1 = 3$$
Two-sided limit: Since both one-sided limits equal 3: $$\lim_{x \to 2} f(x) = 3$$
The limit exists and equals 3. Notice that this is also equal to $f(2) = 2^2 - 1 = 3$, so in this case the function is actually continuous at $x = 2$, despite being defined by different formulas on either side.
Determine whether $\lim_{x \to 0} \frac{|x|}{x}$ exists.
Solution:
Recall that $|x|$ (the absolute value of $x$) equals $x$ when $x > 0$ and equals $-x$ when $x < 0$.
Right-hand limit (approaching 0 from positive values): When $x > 0$, we have $|x| = x$, so: $$\frac{|x|}{x} = \frac{x}{x} = 1$$
Thus $\lim_{x \to 0^+} \frac{|x|}{x} = 1$.
Left-hand limit (approaching 0 from negative values): When $x < 0$, we have $|x| = -x$, so: $$\frac{|x|}{x} = \frac{-x}{x} = -1$$
Thus $\lim_{x \to 0^-} \frac{|x|}{x} = -1$.
Conclusion: The left-hand limit is $-1$ and the right-hand limit is $1$. Since these are not equal:
$$\lim_{x \to 0} \frac{|x|}{x} \text{ does not exist (DNE)}$$
This function has a jump discontinuity at $x = 0$, jumping from $-1$ to $+1$.
Analyze $\lim_{x \to 0} \sin\left(\frac{1}{x}\right)$.
Solution:
This is a fascinating example where the limit fails to exist for a reason different from the previous examples.
As $x$ approaches 0, the quantity $\frac{1}{x}$ grows without bound. When $x = 0.1$, we have $\frac{1}{x} = 10$. When $x = 0.01$, we have $\frac{1}{x} = 100$. When $x = 0.001$, we have $\frac{1}{x} = 1000$.
The sine function oscillates between $-1$ and $1$ as its input increases. So $\sin\left(\frac{1}{x}\right)$ keeps oscillating between $-1$ and $1$ faster and faster as $x$ gets closer to 0.
Let us look at some specific values:
- At $x = \frac{2}{\pi}$, we have $\sin\left(\frac{\pi}{2}\right) = 1$
- At $x = \frac{2}{3\pi}$, we have $\sin\left(\frac{3\pi}{2}\right) = -1$
- At $x = \frac{2}{5\pi}$, we have $\sin\left(\frac{5\pi}{2}\right) = 1$
No matter how close we get to 0, we can always find values of $x$ that make $\sin\left(\frac{1}{x}\right)$ equal to $1$, equal to $-1$, or equal to any value in between.
The function never settles down toward any single value. It oscillates infinitely many times in any neighborhood of 0.
Conclusion: $$\lim_{x \to 0} \sin\left(\frac{1}{x}\right) \text{ does not exist (DNE)}$$
This is not a jump discontinuity (where left and right limits exist but differ) or an infinite discontinuity (where the function heads to infinity). This is oscillatory non-existence: the function simply refuses to approach any particular value.
Find $\lim_{x \to 4} \frac{x^2 - 16}{x - 4}$.
Solution:
If we try direct substitution, we get $\frac{16 - 16}{4 - 4} = \frac{0}{0}$, which is indeterminate. The function is undefined at $x = 4$.
Let us factor the numerator: $$x^2 - 16 = (x - 4)(x + 4)$$
So for $x \neq 4$: $$\frac{x^2 - 16}{x - 4} = \frac{(x - 4)(x + 4)}{x - 4} = x + 4$$
The original function equals $x + 4$ for all $x$ except $x = 4$, where it has a hole.
Since limits only care about what happens near the point (not at the point): $$\lim_{x \to 4} \frac{x^2 - 16}{x - 4} = \lim_{x \to 4} (x + 4) = 8$$
We can verify with a table:
| $x$ | $\frac{x^2 - 16}{x - 4}$ |
|---|---|
| 3.9 | 7.9 |
| 3.99 | 7.99 |
| 4.01 | 8.01 |
| 4.1 | 8.1 |
The values approach 8, confirming our algebraic result.
Key Properties and Rules
Limits of Basic Functions
For polynomial and rational functions (where the denominator is not zero), the limit equals the function value:
$$\lim_{x \to a} p(x) = p(a) \quad \text{for any polynomial } p(x)$$
$$\lim_{x \to a} \frac{p(x)}{q(x)} = \frac{p(a)}{q(a)} \quad \text{if } q(a) \neq 0$$
Limit Laws
If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then:
Sum/Difference: $$\lim_{x \to a} [f(x) \pm g(x)] = L \pm M$$
Constant Multiple: $$\lim_{x \to a} [c \cdot f(x)] = c \cdot L$$
Product: $$\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$$
Quotient: $$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M} \quad \text{provided } M \neq 0$$
Power: $$\lim_{x \to a} [f(x)]^n = L^n$$
These laws let you break complicated limits into simpler pieces.
The Squeeze Theorem (Preview)
If $g(x) \leq f(x) \leq h(x)$ for all $x$ near $a$, and: $$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$$
Then: $$\lim_{x \to a} f(x) = L$$
The function $f(x)$ is “squeezed” between $g(x)$ and $h(x)$, which both approach $L$, so $f(x)$ must also approach $L$. This powerful technique will appear throughout calculus.
Real-World Applications
Instantaneous Speed
This is the application that motivated the entire development of calculus. When you look at your car’s speedometer, it shows your speed at this exact moment, not your average speed over some time interval.
Average speed is easy: distance divided by time. If you drove 60 miles in 2 hours, your average speed was 30 mph. But what about your speed at exactly 1:00:00.000 PM?
Instantaneous speed is the limit of average speeds over shorter and shorter time intervals:
$$\text{instantaneous speed} = \lim_{\Delta t \to 0} \frac{\text{change in position}}{\Delta t}$$
This limit is exactly what a derivative calculates, which is why limits are the foundation of calculus.
Population Growth and Carrying Capacity
When a population grows in an environment with limited resources, it does not grow forever. Instead, it approaches a maximum sustainable population called the carrying capacity.
If $P(t)$ represents population at time $t$, then: $$\lim_{t \to \infty} P(t) = K$$
where $K$ is the carrying capacity. The population gets closer and closer to $K$ but may never exactly reach it. This is limit behavior over time.
Temperature Equilibrium
If you place a hot cup of coffee in a room, it cools down. But it does not cool forever. It approaches room temperature:
$$\lim_{t \to \infty} T(t) = T_{\text{room}}$$
The coffee temperature approaches room temperature asymptotically, getting closer and closer but never quite reaching it (in idealized models). This is Newton’s Law of Cooling in action.
Compound Interest and the Number $e$
Suppose you invest one dollar at 100% annual interest. How much do you have after one year?
- Compounded once: $\$2.00$
- Compounded twice (50% each half-year): $\$2.25$
- Compounded monthly: $\$2.61$
- Compounded daily: $\$2.71$
- Compounded every second: $\$2.718…$
As you compound more frequently, your balance approaches a limit:
$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \approx 2.71828…$$
This is one of the most important numbers in mathematics, and it emerges naturally from a limit. It appears throughout calculus, probability, and the natural sciences.
Self-Test Problems
Problem 1: Use a table to find $\lim_{x \to 2} (x^2 - 3x + 5)$.
Show Answer
Let us build a table approaching $x = 2$ from both sides:
| $x$ | $x^2 - 3x + 5$ |
|---|---|
| 1.9 | $3.61 - 5.7 + 5 = 2.91$ |
| 1.99 | $3.9601 - 5.97 + 5 = 2.9901$ |
| 2.01 | $4.0401 - 6.03 + 5 = 3.0101$ |
| 2.1 | $4.41 - 6.3 + 5 = 3.11$ |
The values approach 3 from both sides.
Alternatively, since this is a polynomial, we can substitute directly: $$(2)^2 - 3(2) + 5 = 4 - 6 + 5 = 3$$
$$\lim_{x \to 2} (x^2 - 3x + 5) = 3$$
Problem 2: Find $\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$.
Show Answer
Direct substitution gives $\frac{0}{0}$, which is indeterminate.
Factor the numerator: $$x^2 - 25 = (x-5)(x+5)$$
For $x \neq 5$: $$\frac{x^2 - 25}{x - 5} = \frac{(x-5)(x+5)}{x-5} = x + 5$$
Therefore: $$\lim_{x \to 5} \frac{x^2 - 25}{x - 5} = \lim_{x \to 5} (x + 5) = 10$$
Problem 3: For the function $f(x) = \begin{cases} 3x - 1 & \text{if } x < 1 \ 5 & \text{if } x = 1 \ 2x + 1 & \text{if } x > 1 \end{cases}$, find $\lim_{x \to 1} f(x)$ or explain why it does not exist.
Show Answer
Left-hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x - 1) = 3(1) - 1 = 2$$
Right-hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x + 1) = 2(1) + 1 = 3$$
Since $2 \neq 3$, the one-sided limits are not equal.
Therefore, $\lim_{x \to 1} f(x)$ does not exist (DNE).
Note: The function value $f(1) = 5$ is irrelevant to whether the limit exists. The limit depends only on behavior near the point, not at the point.
Problem 4: Does $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$ exist? If so, find it.
Show Answer
This is trickier than Example 5 because we now have $x$ multiplied by $\sin\left(\frac{1}{x}\right)$.
We know that $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$ for all $x \neq 0$.
Multiplying by $|x|$: $$-|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x|$$
(The inequality direction may flip depending on the sign of $x$, but the absolute value bounds always work.)
As $x \to 0$:
- $\lim_{x \to 0} (-|x|) = 0$
- $\lim_{x \to 0} |x| = 0$
By the Squeeze Theorem, since $x \sin\left(\frac{1}{x}\right)$ is trapped between two functions that both approach 0:
$$\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$$
The limit exists and equals 0. The $x$ factor “tames” the wild oscillations of $\sin\left(\frac{1}{x}\right)$.
Problem 5: Find $\lim_{x \to 3^+} \frac{x + 2}{x - 3}$ and $\lim_{x \to 3^-} \frac{x + 2}{x - 3}$.
Show Answer
As $x \to 3$, the numerator approaches $3 + 2 = 5$ (a positive number).
The denominator approaches 0, but we need to determine whether it approaches 0 from positive or negative values.
Right-hand limit ($x \to 3^+$): When $x > 3$, the denominator $x - 3 > 0$ (positive). So we have a positive numerator divided by a small positive denominator: $$\lim_{x \to 3^+} \frac{x + 2}{x - 3} = +\infty$$
Left-hand limit ($x \to 3^-$): When $x < 3$, the denominator $x - 3 < 0$ (negative). So we have a positive numerator divided by a small negative denominator: $$\lim_{x \to 3^-} \frac{x + 2}{x - 3} = -\infty$$
Since the one-sided limits are not equal (and not finite), the two-sided limit $\lim_{x \to 3} \frac{x + 2}{x - 3}$ does not exist. There is a vertical asymptote at $x = 3$.
Summary
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A limit describes the value a function approaches as its input approaches some target, written as $\lim_{x \to a} f(x) = L$.
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Limits are about behavior near a point, not at the point. A function can have a limit at a point where it is undefined.
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One-sided limits describe approach from a single direction: $\lim_{x \to a^-} f(x)$ (from the left) and $\lim_{x \to a^+} f(x)$ (from the right).
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A two-sided limit exists only if both one-sided limits exist and are equal.
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Limits can fail to exist for several reasons: jump discontinuities (different one-sided limits), infinite behavior (function heads to $\pm\infty$), or oscillating behavior (function never settles down).
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You can find limits using tables (calculating values closer and closer to the target), graphs (tracing where the curve heads), or algebra (simplifying expressions).
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Limits appear throughout the real world: instantaneous velocity, population growth approaching carrying capacity, temperatures approaching equilibrium, and compound interest approaching $e$.
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Limits are the foundation of calculus. Derivatives and integrals are both defined using limits, making this concept essential for everything that follows.
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The key insight: limits let us make rigorous sense of “infinitely close” and “instantaneous” without ever having to divide by zero or deal with actual infinities.