1. Why limits exist as a concept
Consider the function
$$ f(x) = \frac{x^2 - 1}{x - 1}. $$What is $f(1)$? Plugging in gives $0/0$ — undefined. The function refuses to answer. But look at what $f$ does for values near $1$:
| $x$ | $0.9$ | $0.99$ | $0.999$ | $1$ | $1.001$ | $1.01$ | $1.1$ |
|---|---|---|---|---|---|---|---|
| $f(x)$ | $1.9$ | $1.99$ | $1.999$ | undef. | $2.001$ | $2.01$ | $2.1$ |
The function is undefined at $x = 1$, but every nearby value of $x$ produces an output close to $2$. The closer you get to $1$, the closer $f(x)$ gets to $2$. There's a value the function is aiming at, even though it never reaches it. That value is the limit.
Limits are how mathematics talks about "what would the answer be if we could get there" — for points where the function is undefined, jumps, or behaves badly. Without this concept, you can't define a derivative, a tangent line, an instantaneous rate, or what it means for a function to be smooth.
2. The informal definition
We write
$$ \lim_{x \to a} f(x) = L $$to mean: as $x$ gets arbitrarily close to $a$ (from either side, but $x \neq a$), the value $f(x)$ gets arbitrarily close to $L$.
Three things this definition is not saying:
- It is not saying $f(a) = L$. The function might not be defined at $a$ at all, or it might be defined but equal to something else. The limit ignores the single point $x = a$ entirely.
- It is not saying $x$ ever equals $a$. The phrase "$x \to a$" means $x$ approaches $a$, and the inequality $x \neq a$ is baked in.
- It is not a recipe for computing anything yet. It's a meaning. The recipes come later in the page.
You'll also see $f(x) \to L$ as $x \to a$, which is the same statement read in arrow form. The limit symbol $\lim$ and the arrow $\to$ are interchangeable; pick whichever reads more naturally for the sentence you're writing.
3. Reading a limit from a table
The numerical approach is exactly what we just did with $f(x) = (x^2 - 1)/(x - 1)$: pick a sequence of $x$-values that closes in on $a$ from both sides, evaluate $f$, and watch where the outputs settle. Used carefully, a table is a perfectly valid way to guess a limit — and it's the surest way to build intuition for what the symbol means.
Try $\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$ (with $x$ in radians):
| $x$ | $-0.1$ | $-0.01$ | $-0.001$ | $0$ | $0.001$ | $0.01$ | $0.1$ |
|---|---|---|---|---|---|---|---|
| $\sin(x)/x$ | $0.99833$ | $0.99998$ | $1.00000$ | undef. | $1.00000$ | $0.99998$ | $0.99833$ |
The outputs squeeze in on $1$ from both sides. The limit is $1$. This particular limit is the cornerstone of calculus — it's why the derivative of $\sin x$ is $\cos x$ — and you can't prove it algebraically. A table is the cleanest way to see it for the first time.
Tables can mislead. The function $\sin(\pi/x)$ near $x = 0$ produces $0$ at $x = 1, \tfrac{1}{2}, \tfrac{1}{3}, \dots$ but oscillates wildly between them — a sparse table will suggest a limit of $0$ when there is no limit at all. Numerical evidence is a strong hint, not a proof.
4. Reading a limit from a graph
Graphically, $\lim_{x \to a} f(x) = L$ means: as you walk along the curve toward $x = a$ from either side, your height approaches $L$. Whether the curve actually touches that height at $a$ — or has a hole, or has a different value — doesn't matter for the limit.
Here is $f(x) = (x^2 - 1)/(x - 1)$, which agrees with $y = x + 1$ everywhere except at $x = 1$, where it has a hole:
The hole is the function's way of saying "I can't answer here." The limit is mathematics' way of saying "but the answer it would have given is $2$." That distinction is the entire reason limits matter.
5. One-sided limits
Sometimes a function approaches different values depending on which side you come from. To talk about this, we split the limit in two:
- $\displaystyle\lim_{x \to a^-} f(x)$ — the left-hand limit: $x$ approaches $a$ through values less than $a$.
- $\displaystyle\lim_{x \to a^+} f(x)$ — the right-hand limit: $x$ approaches $a$ through values greater than $a$.
The two-sided limit $\displaystyle\lim_{x \to a} f(x)$ exists if and only if both one-sided limits exist and agree. If they disagree, the two-sided limit does not exist (often written DNE).
Consider the step-style function
$$ g(x) = \begin{cases} x + 1 & x < 0, \\ x^2 & x \ge 0. \end{cases} $$From the left, $g(x) \to 0 + 1 = 1$. From the right, $g(x) \to 0^2 = 0$. The left limit is $1$, the right limit is $0$, they disagree, and $\lim_{x \to 0} g(x)$ doesn't exist. The graph has a jump at $x = 0$.
A two-sided limit existing means the two one-sided journeys arrive at the same place. A removable discontinuity (a hole) still has a two-sided limit — both sides agree on a value the function happens to miss. A jump has different one-sided limits and so no two-sided limit at all.
6. Limit laws
Limits compose with arithmetic exactly the way you would hope. Assume $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, both existing as finite numbers. Then:
| Operation | Law |
|---|---|
| Sum | $\displaystyle\lim_{x \to a}\bigl[f(x) + g(x)\bigr] = L + M$ |
| Difference | $\displaystyle\lim_{x \to a}\bigl[f(x) - g(x)\bigr] = L - M$ |
| Constant multiple | $\displaystyle\lim_{x \to a}\bigl[c \cdot f(x)\bigr] = c \cdot L$ |
| Product | $\displaystyle\lim_{x \to a}\bigl[f(x) \cdot g(x)\bigr] = L \cdot M$ |
| Quotient | $\displaystyle\lim_{x \to a}\dfrac{f(x)}{g(x)} = \dfrac{L}{M}$ (provided $M \neq 0$) |
| Power | $\displaystyle\lim_{x \to a}\bigl[f(x)\bigr]^n = L^n$ |
These are stated here — not proved. The proofs use the $\varepsilon$–$\delta$ machinery you'll meet in calculus, and they're routine once you have it. For now, treat the laws as licences: if you can find the limit of each piece, the limit of the combination behaves the way arithmetic predicts.
Combined with two anchor facts — $\lim_{x \to a} c = c$ for any constant, and $\lim_{x \to a} x = a$ — the laws let you take limits of every polynomial by just plugging in.
7. Direct substitution
The single most important computational shortcut: if $f$ is continuous at $a$, then
$$ \lim_{x \to a} f(x) = f(a). $$That is, you can find the limit by just plugging in. Continuous functions include every polynomial, every rational function (away from points where the denominator is zero), the trig functions in their natural domains, exponentials, logarithms, and any sum, product, quotient, or composition of these.
For now, the working definition of "continuous at $a$" is the intuitive one: the graph passes through $(a, f(a))$ without lifting your pencil. The rigorous three-condition definition (function defined, limit exists, the two are equal) is calculus territory — see the linked Calculus · Limits page.
Some examples worked entirely by substitution:
- $\displaystyle\lim_{x \to 2}(3x + 1) = 3(2) + 1 = 7$.
- $\displaystyle\lim_{x \to 3}(x^2 - 2x + 5) = 9 - 6 + 5 = 8$.
- $\displaystyle\lim_{x \to 1}\frac{x^2 + 1}{x + 2} = \frac{2}{3}$.
That handles the easy half of every limit problem you'll see in precalculus. The interesting half is what happens when substitution gives you $0/0$.
8. The $0/0$ indeterminate form
Suppose substitution gives
$$ \lim_{x \to a} \frac{f(x)}{g(x)} \;\;\text{looks like}\;\; \frac{0}{0}. $$This is called an indeterminate form. It does not mean the limit is undefined — it means the form you're looking at doesn't determine the answer. The limit might be $0$, or $7$, or $-\infty$, or DNE. You have to do more work to find out.
The two precalculus techniques are factor and cancel and rationalize. Each one removes the offending zero in the denominator algebraically, leaving a form you can substitute into.
Factor and cancel
If both top and bottom are polynomials and both vanish at $a$, then $(x - a)$ is a factor of each — you can factor it out, cancel, and plug in.
$$ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \;=\; \lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} \;=\; \lim_{x \to 3} (x + 3) \;=\; 6. $$The expressions $(x - 3)(x + 3)/(x - 3)$ and $x + 3$ disagree at exactly one point: $x = 3$. The limit ignores that one point. So as far as the limit is concerned, the two expressions are the same.
Rationalize
When a square root is causing the $0/0$, multiply top and bottom by the conjugate — the same expression with the sign in the middle flipped. The difference-of-squares identity turns the radical into a polynomial.
$$ \lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \;=\; \lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2} \;=\; \lim_{x \to 0} \frac{(x + 4) - 4}{x(\sqrt{x + 4} + 2)} $$The numerator simplifies to $x$, which cancels the $x$ in the denominator:
$$ \;=\; \lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2} \;=\; \frac{1}{\sqrt{4} + 2} \;=\; \frac{1}{4}. $$$\lim_{x \to 0} \sin(x)/x$ is $0/0$ but neither factoring nor rationalizing helps — there's no algebraic relationship between $\sin x$ and $x$ to exploit. This one needs the squeeze theorem (or a power series), which is calculus material. Some limits are genuinely beyond precalc's toolkit, and that's fine.
9. Limits at infinity
The notation
$$ \lim_{x \to \infty} f(x) = L $$means: as $x$ grows without bound, $f(x)$ approaches $L$. Graphically, $y = L$ is a horizontal asymptote — the curve flattens toward that height as it runs off to the right. The mirror statement $x \to -\infty$ describes behaviour as $x$ runs off to the left.
For rational functions $p(x)/q(x)$, the answer is decided by the leading terms — the highest-power terms in numerator and denominator. Every other term becomes negligible at very large $x$.
| Comparison | Limit at $\infty$ | Example |
|---|---|---|
| $\deg(p) < \deg(q)$ | $0$ | $\displaystyle\lim_{x \to \infty} \frac{2x + 1}{x^2} = 0$ |
| $\deg(p) = \deg(q)$ | ratio of leading coefficients | $\displaystyle\lim_{x \to \infty} \frac{3x^2 + 1}{x^2 + x} = 3$ |
| $\deg(p) > \deg(q)$ | $\pm\infty$ (diverges) | $\displaystyle\lim_{x \to \infty} \frac{x^3 + 1}{x + 5} = \infty$ |
The mechanical justification: divide every term, top and bottom, by the highest power of $x$ that appears in the denominator. The "junk" terms become $1/x$, $1/x^2$, etc. — each tending to $0$ as $x \to \infty$ — and the answer falls out.
Example, equal degrees:
$$ \lim_{x \to \infty} \frac{3x^2 + 1}{x^2 + x} \;=\; \lim_{x \to \infty} \frac{3 + 1/x^2}{1 + 1/x} \;=\; \frac{3 + 0}{1 + 0} \;=\; 3. $$Strictly, "$\lim = \infty$" is shorthand. Infinity is not a real number, so the limit doesn't "exist" as a finite value — but the notation captures the precise way in which $f$ grows without bound, and treating it as a limit is standard practice.
10. Common pitfalls
This is true only when $f$ is continuous at $a$. The whole point of limits is to describe behaviour near $a$ for functions where $f(a)$ is undefined, different, or pathological. The limit and the value are two separate concepts that happen to coincide on nice functions.
$0/0$ is indeterminate, not undefined. It's a flag that more work is needed — factor, rationalize, or (in calculus) apply L'Hôpital. Indeterminate forms hide every limit that matters.
If $\lim_{x \to a^-}$ and $\lim_{x \to a^+}$ are different numbers, the two-sided limit does not exist. You don't get to average them or pick one. The two-sided limit only exists when both one-sided journeys end at the same place.
When you cancel $(x - a)$ from top and bottom, you're computing the limit of a function that has a removable hole at $x = a$. The cancelled expression and the original are not the same function — they agree everywhere except at $a$. The limit is the same, but the functions themselves differ. Important when you later want to describe continuity or the graph.
The dominant term is the one with the highest power, not the one with the largest coefficient. In $\dfrac{1000x + 1}{x^2}$ the $x^2$ wins by a mile at large $x$, and the limit is $0$ — not $1000$.
11. Worked examples
Try each one before opening the solution. The aim is to recognize which technique a limit calls for as quickly as you can read the expression.
Example 1 · Direct substitution: $\displaystyle\lim_{x \to 4} \sqrt{2x + 1}$
The function $\sqrt{2x + 1}$ is continuous wherever $2x + 1 \ge 0$, which includes $x = 4$. Substitute:
$$ \lim_{x \to 4} \sqrt{2x + 1} = \sqrt{2(4) + 1} = \sqrt{9} = 3. $$Example 2 · Factor and cancel: $\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$
Step 1. Substitution gives $0/0$ — indeterminate.
Step 2. Factor the numerator as a difference of squares:
$$ \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2 \quad (\text{for } x \neq 2). $$Step 3. Take the limit of the simplified form:
$$ \lim_{x \to 2}(x + 2) = 4. $$The original function is undefined at $x = 2$, but the limit reaches that height through the hole.
Example 3 · Rationalize: $\displaystyle\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$
Step 1. Substitution gives $0/0$.
Step 2. Multiply top and bottom by the conjugate $\sqrt{x} + 3$:
$$ \frac{\sqrt{x} - 3}{x - 9} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \frac{x - 9}{(x - 9)(\sqrt{x} + 3)}. $$Step 3. Cancel $(x - 9)$:
$$ = \frac{1}{\sqrt{x} + 3}. $$Step 4. Substitute $x = 9$:
$$ \lim_{x \to 9} \frac{1}{\sqrt{x} + 3} = \frac{1}{3 + 3} = \frac{1}{6}. $$Example 4 · One-sided limits: $f(x) = \begin{cases} 2x + 1 & x < 1 \\ x^2 & x \ge 1 \end{cases}$ at $x = 1$
Left limit. Use the $2x + 1$ branch:
$$ \lim_{x \to 1^-} f(x) = 2(1) + 1 = 3. $$Right limit. Use the $x^2$ branch:
$$ \lim_{x \to 1^+} f(x) = 1^2 = 1. $$Conclusion. The one-sided limits are $3$ and $1$ — different. So $\lim_{x \to 1} f(x)$ does not exist; the graph jumps at $x = 1$.
Example 5 · Limit at infinity: $\displaystyle\lim_{x \to \infty} \frac{4x^3 - x + 7}{2x^3 + 5x^2}$
Step 1. Numerator and denominator both have degree $3$ — same degree, so the limit is the ratio of leading coefficients.
Step 2. Leading coefficients: $4$ on top, $2$ on bottom.
$$ \lim_{x \to \infty} \frac{4x^3 - x + 7}{2x^3 + 5x^2} = \frac{4}{2} = 2. $$If you want to see the mechanics: divide every term by $x^3$. The fractions $1/x^2$, $7/x^3$, $5/x$ all tend to $0$, leaving $4/2 = 2$.
Example 6 · Limit at infinity with degree mismatch: $\displaystyle\lim_{x \to \infty} \frac{2x + 1}{x^2 + 3}$
The denominator's degree ($2$) exceeds the numerator's ($1$), so the denominator grows faster and the quotient is squeezed toward $0$.
$$ \lim_{x \to \infty} \frac{2x + 1}{x^2 + 3} = 0. $$Mechanical check: divide top and bottom by $x^2$ — $\dfrac{2/x + 1/x^2}{1 + 3/x^2} \to \dfrac{0 + 0}{1 + 0} = 0$.
12. What's next: $\varepsilon$–$\delta$
Everything on this page has been built on the phrase "arbitrarily close." It's a useful intuition, but it's slippery — what does "arbitrarily close" actually mean as a logical statement? Could two people disagree on whether a sequence of outputs is "close enough" to $L$?
Calculus replaces the phrase with a precise game: $\lim_{x \to a} f(x) = L$ if, for any tolerance $\varepsilon > 0$ you demand on the output, there exists a tolerance $\delta > 0$ on the input such that $0 < |x - a| < \delta$ guarantees $|f(x) - L| < \varepsilon$. This is the $\varepsilon$–$\delta$ definition, due to Weierstrass — and it's what the Calculus · Limits page builds out in full.
For now: trust the intuition, get fluent with the techniques, and the formalism will feel like a natural sharpening rather than a foreign object when you meet it.