1. The big idea: approaching without arriving
Imagine walking toward a wall. With each step you halve your remaining distance: one metre, then half a metre, then a quarter, then an eighth. You will never touch the wall — but it is unambiguous which wall you are walking toward. That target — the value you are heading for, regardless of whether you ever get there — is what a limit captures.
We say $\lim_{x \to a} f(x) = L$ if the values of $f(x)$ can be made arbitrarily close to $L$ by choosing $x$ sufficiently close to $a$ — but with $x \neq a$.
Read the notation aloud: "the limit of $f$ of $x$, as $x$ approaches $a$, equals $L$". Every word matters. The variable $x$ is approaching $a$ — never arriving. The output $f(x)$ is approaching $L$. And the equation is a statement about what happens near $a$, not at $a$.
That last clause is the entire point. The value $f(a)$ might be $L$, or some other number, or simply undefined. The limit doesn't care. Whatever the function is doing at $a$ itself, the limit asks a different question: where is the function headed?
The arrow $x \to a$ is read "approaches" or "tends to". Writing $x \to a$ silently asserts $x \neq a$ — the limit is always evaluated nearby, never at, the target point.
2. A motivating example: the hole in the graph
Consider the function
$$ f(x) = \frac{x^2 - 1}{x - 1}. $$At $x = 1$ this function is undefined — the denominator is zero, and division by zero is not a number. But what happens to $f(x)$ as $x$ approaches $1$? Try a few values from each side:
| $x$ (from the left) | $f(x)$ | $x$ (from the right) | $f(x)$ |
|---|---|---|---|
| $0.9$ | $1.9$ | $1.1$ | $2.1$ |
| $0.99$ | $1.99$ | $1.01$ | $2.01$ |
| $0.999$ | $1.999$ | $1.001$ | $2.001$ |
| $0.9999$ | $1.9999$ | $1.0001$ | $2.0001$ |
From both sides, $f(x)$ is closing in on $2$ — and we can get as close as we like by squeezing $x$ closer to $1$. The function is not defined at $x = 1$, but the values it takes near $x = 1$ tell a completely consistent story.
$$ \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2. $$The reason for this is hiding in plain sight. For every $x \neq 1$ we may factor and cancel:
$$ \frac{x^2 - 1}{x - 1} \;=\; \frac{(x - 1)(x + 1)}{x - 1} \;=\; x + 1. $$So everywhere except at $x = 1$, our function is identical to the line $y = x + 1$. The graph looks like that line — with a single point missing, a "hole" right at $(1, 2)$. The limit is asking: what would the value have to be to fill that hole?
$f(1)$ does not exist. $\lim_{x \to 1} f(x)$ does exist, and equals $2$. These are two separate questions, and one of the most important habits in calculus is learning to keep them apart.
3. One-sided limits
So far we have implicitly checked that $f(x)$ approaches the same value whether $x$ creeps toward $a$ from below or from above. That equality is a real condition — and when it fails, the limit fails too.
A one-sided limit only looks at one direction of approach:
- $\displaystyle\lim_{x \to a^-} f(x)$ — 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 is built from these two pieces, and exists only when they agree:
$$ \lim_{x \to a} f(x) = L \quad\Longleftrightarrow\quad \lim_{x \to a^-} f(x) = L \;\text{ and }\; \lim_{x \to a^+} f(x) = L. $$Classic example: the sign function $\operatorname{sgn}(x)$, which is $-1$ for $x < 0$, $0$ at $x = 0$, and $+1$ for $x > 0$. As $x \to 0^-$ it stays at $-1$; as $x \to 0^+$ it stays at $+1$. Both one-sided limits exist — but they disagree, so the two-sided limit $\lim_{x \to 0} \operatorname{sgn}(x)$ does not exist.
"The limit does not exist" is a perfectly valid answer in calculus — it is not a sign you have made a mistake. It is the function telling you it has no single target as it approaches the point.
4. Computing limits
In practice you rarely build a table and stare at it. Three techniques will handle the great majority of limits you meet.
4.1 Direct substitution
If $f$ is continuous at $a$, the limit is just the value:
$$ \lim_{x \to a} f(x) = f(a). $$Polynomials are continuous everywhere; rational functions are continuous wherever their denominators are nonzero; $\sin, \cos, e^x$ are continuous everywhere. So when you see a "nice" expression and the substitution doesn't blow up, you are done. For instance,
$$ \lim_{x \to 2} \left( 3x^2 - x + 4 \right) \;=\; 3(2)^2 - 2 + 4 \;=\; 14. $$4.2 Algebraic simplification — taming "0/0"
When direct substitution produces something nonsensical like $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$, the limit isn't necessarily undefined — those expressions are called indeterminate forms because the symbol alone doesn't tell you the answer. The actual value depends on how the numerator and denominator are going to zero (or infinity).
The most common rescue is to factor, cancel a common term, and try again. We already did this with $\tfrac{x^2 - 1}{x - 1}$. Here is a sharper case:
$$ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \;=\; \lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} \;=\; \lim_{x \to 3} (x + 3) \;=\; 6. $$The factor $x - 3$ is the source of the trouble; cancelling it removes the trouble; then substitution finishes the job.
4.3 Standard limits worth knowing
Some limits come up so often that you should recognise them on sight rather than re-derive them. A few essentials:
$$ \begin{aligned} \lim_{x \to 0} \frac{\sin x}{x} &= 1, \\[4pt] \lim_{x \to 0} \frac{1 - \cos x}{x} &= 0, \\[4pt] \lim_{x \to 0} \frac{e^x - 1}{x} &= 1, \\[4pt] \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{\!n} &= e. \end{aligned} $$The first one is the workhorse of trigonometric calculus — it is what makes the derivative of $\sin$ come out to $\cos$.
Always try substitution first. If it works, stop. If it produces an indeterminate form, look for an algebraic move (factoring, conjugates, common denominators). If the expression already looks like a recognized standard limit, use that. Reach for fancier tools (L'Hôpital, series) only when these three fail.
5. Limits at infinity
So far the input $x$ has been approaching a finite point $a$. We can also ask: what does $f(x)$ do as $x$ grows without bound?
$$ \lim_{x \to \infty} f(x) \;=\; L $$means that $f(x)$ can be made arbitrarily close to $L$ by taking $x$ large enough. Geometrically, the graph has a horizontal asymptote at $y = L$.
For rational functions $\tfrac{p(x)}{q(x)}$, there is a quick rule. Compare the degree of the numerator $p$ with the degree of the denominator $q$:
| Relationship | $\displaystyle\lim_{x \to \infty} \tfrac{p(x)}{q(x)}$ | Reason |
|---|---|---|
| $\deg p < \deg q$ | $0$ | Denominator outruns numerator. |
| $\deg p = \deg q$ | ratio of leading coefficients | Both grow at the same rate; only the head terms matter. |
| $\deg p > \deg q$ | $\pm \infty$ | Numerator outruns denominator; the limit doesn't exist as a finite number. |
For example,
$$ \lim_{x \to \infty} \frac{3x^2 + 5x - 1}{7x^2 - 2x + 4} \;=\; \frac{3}{7}. $$Only the leading $3x^2$ and $7x^2$ end up mattering. The lower-order terms become negligible compared to the leading terms — a fact you can confirm by dividing numerator and denominator by $x^2$ and taking the limit term by term.
Analogously, $\lim_{x \to -\infty}$ asks where the function is heading as $x$ marches off in the negative direction. The same degree rule applies; signs may differ.
6. Why limits matter
Limits are not a self-contained party trick. They are the foundation on which the rest of calculus stands — without them, the two big ideas of the subject have no rigorous meaning.
The derivative of $f$ at $a$ is the slope of the tangent line at that point. But the tangent line is the limiting position of secant lines as the second point slides toward the first. It is defined by
$$ f'(a) \;=\; \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}. $$Without the limit, the quotient is $\tfrac{0}{0}$ at $h = 0$ — formally undefined, just like our motivating example. The whole reason derivatives exist as numbers is that this limit, in the situations where calculus applies, has a specific finite value.
The definite integral $\int_a^b f(x)\,dx$ is the area under a curve. But the "area under a curve" is itself the limit of Riemann sums — approximations by rectangles whose number grows without bound while each rectangle shrinks:
$$ \int_a^b f(x)\,dx \;=\; \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \, \Delta x. $$Again, every individual sum on the right is a finite, perfectly defined quantity. The integral is what those finite sums approach.
Calculus is the mathematics of change. Limits are the mathematics that makes "infinitely small change" a well-defined idea instead of a paradox.
Everything else in the subject — continuity, differentiability, infinite series, Taylor expansions, improper integrals — is built on top of this single move. Get comfortable with limits, and the rest of calculus opens up. Stay uneasy with them, and every subsequent definition will feel like it is balanced on something you don't quite trust.
7. Going further: $\varepsilon$–$\delta$, L'Hôpital, squeeze, continuity
Everything so far has been the operational picture — enough to compute almost any limit that appears in a first-year course. The next layer is what makes those operations justifiable: a precise definition, a couple of powerful theorems, and the way limits glue into the formal definition of continuity.
7.1 The formal $\varepsilon$–$\delta$ definition
The informal "arbitrarily close" and "sufficiently close" can be sharpened into a single statement, due to Cauchy and Weierstrass, that has anchored real analysis for over a century.
$\displaystyle \lim_{x \to a} f(x) = L$ means: for every $\varepsilon > 0$, there exists a $\delta > 0$ such that whenever $0 < |x - a| < \delta$, we have $|f(x) - L| < \varepsilon$.
Read as a game: you challenge me with a tolerance $\varepsilon$ — "get $f(x)$ within $\varepsilon$ of $L$" — and I must reply with a radius $\delta$ around $a$ inside which the function honours that tolerance. If I can always answer, no matter how small you make $\varepsilon$, the limit is $L$.
The condition $0 < |x - a|$ is the formal way of saying "$x \neq a$" — the limit ignores the value at $a$ itself, exactly as the informal definition promised.
7.2 Limit laws
Once a limit exists, it behaves algebraically the way you would hope. If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then:
$$ \begin{aligned} \lim_{x \to a} \big( f(x) \pm g(x) \big) &= L \pm M, \\[4pt] \lim_{x \to a} \big( f(x) \cdot g(x) \big) &= L \cdot M, \\[4pt] \lim_{x \to a} \frac{f(x)}{g(x)} &= \frac{L}{M} \quad (M \neq 0), \\[4pt] \lim_{x \to a} \big( c \cdot f(x) \big) &= c \cdot L. \end{aligned} $$These are the silent backbone of direct substitution: a polynomial's limit is computed term-by-term because the sum, product, and constant-multiple laws say it can be.
7.3 The full menagerie of indeterminate forms
Direct substitution can fail in seven distinct ways. Each is a signal — not a verdict — that you need to do more work.
| Form | Typical setting | Standard rescue |
|---|---|---|
| $\tfrac{0}{0}$ | $\tfrac{x^2 - 1}{x - 1}$ as $x \to 1$ | Factor & cancel; rationalise; L'Hôpital |
| $\tfrac{\infty}{\infty}$ | $\tfrac{3x^2 + x}{7x^2 - 5}$ as $x \to \infty$ | Divide by highest power; L'Hôpital |
| $0 \cdot \infty$ | $x \ln x$ as $x \to 0^+$ | Rewrite as $\tfrac{\ln x}{1/x}$, then $\tfrac{\infty}{\infty}$ |
| $\infty - \infty$ | $\sqrt{x^2 + x} - x$ as $x \to \infty$ | Combine into a single fraction or rationalise |
| $1^{\infty}$ | $(1 + 1/n)^n$ as $n \to \infty$ | Take $\ln$, evaluate, exponentiate |
| $0^0$ | $x^x$ as $x \to 0^+$ | Same — $\ln$, evaluate, exponentiate |
| $\infty^0$ | $x^{1/x}$ as $x \to \infty$ | Same — $\ln$, evaluate, exponentiate |
Notice the pattern: only $\tfrac{0}{0}$ and $\tfrac{\infty}{\infty}$ are the "primary" forms. Everything else can be massaged into one of those two and then attacked with the same tools. By contrast, expressions like $\tfrac{1}{0}$, $\infty + \infty$, or $0^{\infty}$ are not indeterminate — they have unambiguous behaviour ($\pm \infty$, $\infty$, and $0$ respectively).
7.4 L'Hôpital's rule
When you hit $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ and no algebraic simplification leaps out, L'Hôpital's rule converts the limit of a ratio into the limit of the ratio of derivatives.
If $\lim_{x \to a} \tfrac{f(x)}{g(x)}$ is of the form $\tfrac{0}{0}$ or $\tfrac{\pm\infty}{\pm\infty}$, and $f$ and $g$ are differentiable near $a$ (with $g'(x) \neq 0$ nearby), then
$$ \lim_{x \to a} \frac{f(x)}{g(x)} \;=\; \lim_{x \to a} \frac{f'(x)}{g'(x)}, $$provided the right-hand limit exists (or is $\pm\infty$). The same statement holds for one-sided limits and for $a = \pm\infty$.
For example, the standard limit $\lim_{x \to 0} \tfrac{\sin x}{x}$ is of the form $\tfrac{0}{0}$, and L'Hôpital gives
$$ \lim_{x \to 0} \frac{\sin x}{x} \;=\; \lim_{x \to 0} \frac{\cos x}{1} \;=\; 1. $$Two cautions. First, the rule only applies to $\tfrac{0}{0}$ and $\tfrac{\infty}{\infty}$ — applying it to a quotient that is not indeterminate (say $\tfrac{0}{1}$) usually produces the wrong answer. Always verify the form first. Second, you differentiate the numerator and denominator separately; this is not the quotient rule.
7.5 The squeeze theorem
Some limits are hard to compute directly but easy to corner. If you can trap $f(x)$ between two functions whose limits agree at $a$, the trapped function must agree too.
If $g(x) \le f(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$ itself), and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$ as well.
The classic application is $\lim_{x \to 0} x^2 \sin(1/x)$. The factor $\sin(1/x)$ oscillates wildly as $x \to 0$ — it has no limit on its own — but it is bounded between $-1$ and $1$, so
$$ -x^2 \;\le\; x^2 \sin\!\tfrac{1}{x} \;\le\; x^2. $$Both bounding functions tend to $0$ as $x \to 0$, so the wild middle is squeezed to $0$ along with them.
7.6 Infinite limits and vertical asymptotes
Distinct from "limits at infinity" — where the input grows — are infinite limits, where the output grows without bound as $x$ approaches a finite point. Writing $\lim_{x \to a} f(x) = +\infty$ means $f(x)$ exceeds any chosen bound once $x$ is close enough to $a$. The graph has a vertical asymptote at $x = a$.
The cleanest example is $f(x) = 1/x^2$. As $x \to 0$ from either side, $1/x^2$ grows without bound:
$$ \lim_{x \to 0} \frac{1}{x^2} \;=\; +\infty. $$Compare $f(x) = 1/x$. From the right, $1/x \to +\infty$; from the left, $1/x \to -\infty$. The two one-sided infinite limits disagree, so the two-sided $\lim_{x \to 0} 1/x$ does not exist (not even in the extended sense).
7.7 Continuity, defined through limits
Continuity is exactly the condition that the limit and the function agree — formally:
A function $f$ is continuous at $a$ when all three hold:
- $f(a)$ is defined,
- $\lim_{x \to a} f(x)$ exists,
- $\lim_{x \to a} f(x) = f(a)$.
If any of the three fails, $f$ has a discontinuity at $a$:
- Removable — the limit exists but disagrees with $f(a)$ (or $f(a)$ is undefined). The hole in $\tfrac{x^2-1}{x-1}$ at $x = 1$ is the canonical example; redefining $f(1) = 2$ heals it.
- Jump — the one-sided limits exist but disagree. The sign function at $0$ is a jump.
- Infinite — at least one one-sided limit is $\pm\infty$. $1/x$ at $0$ is an infinite (essential) discontinuity.
This is why direct substitution works: a function is continuous exactly when "plug in the number" gives the right answer. Every named family — polynomials, rationals on their domain, $\sin$, $\cos$, $e^x$, $\ln$ on its domain — is continuous wherever it is defined, which is why so many limits are answered in one step.
8. Common pitfalls
The value of a function at a point and the limit of the function at that point are different questions with different answers. A function need not equal its limit at the point — it may be undefined there, or take some unrelated value (a "removable discontinuity"). When the two do agree, we say $f$ is continuous at $a$; that is a special property, not an automatic one.
A two-sided limit fails the moment the left and right approaches disagree. Always check both sides for piecewise functions, absolute values, and anything involving square roots or denominators that change sign across the point. "It works from the right" is half an answer.
$\tfrac{0}{0}$ is called indeterminate for a reason — the value of the limit depends on the specific functions involved, not on the symbol itself. You'll see $\tfrac{0}{0}$ produce $2$, or $6$, or $\infty$, or anything you like. Treat it as a signal to do more work (factor, simplify, recognise a standard form), not as a verdict.
$\infty$ is shorthand for "grows without bound", not a quantity you can add, subtract, or substitute. Writing $\lim = \infty$ means "the limit fails to exist, in this particular runaway way" — useful information, but it is not an equation between numbers. Statements like $\infty - \infty = 0$ are nonsense; they are also indeterminate forms.
9. Worked examples
Open each one only after you have tried it yourself. The goal is to recognise which technique to reach for, not to memorise specific answers.
Example 1 · Direct substitution: $\displaystyle\lim_{x \to 2} (x^3 - 2x + 5)$
The expression is a polynomial, so it is continuous everywhere. Substitute directly:
$$ \lim_{x \to 2} (x^3 - 2x + 5) \;=\; (2)^3 - 2(2) + 5 \;=\; 8 - 4 + 5 \;=\; 9. $$When substitution lands on a well-defined number, you are done.
Example 2 · Factor and cancel: $\displaystyle\lim_{x \to -2} \frac{x^2 + 5x + 6}{x + 2}$
Direct substitution gives $\tfrac{0}{0}$ — indeterminate. Factor the numerator:
$$ x^2 + 5x + 6 \;=\; (x + 2)(x + 3). $$The shared $(x + 2)$ factor cancels (legal because we are taking $x \to -2$ with $x \neq -2$):
$$ \lim_{x \to -2} \frac{(x + 2)(x + 3)}{x + 2} \;=\; \lim_{x \to -2} (x + 3) \;=\; 1. $$The original function had a removable discontinuity at $x = -2$; the limit fills in the hole.
Example 3 · One-sided check: $\displaystyle\lim_{x \to 0} \frac{|x|}{x}$
Recall $|x| = x$ for $x > 0$ and $|x| = -x$ for $x < 0$.
Right-hand limit. For $x > 0$, $\tfrac{|x|}{x} = \tfrac{x}{x} = 1$, so
$$ \lim_{x \to 0^+} \frac{|x|}{x} \;=\; 1. $$Left-hand limit. For $x < 0$, $\tfrac{|x|}{x} = \tfrac{-x}{x} = -1$, so
$$ \lim_{x \to 0^-} \frac{|x|}{x} \;=\; -1. $$The two one-sided limits disagree, so the two-sided limit does not exist. Both pieces of information matter — "doesn't exist, because the sides disagree, and here they are" is a complete answer.
Example 4 · Limit at infinity of a rational function: $\displaystyle\lim_{x \to \infty} \frac{4x^3 - x + 2}{2x^3 + 7x^2 - 5}$
Numerator and denominator both have degree $3$. By the degree rule, the limit is the ratio of leading coefficients:
$$ \lim_{x \to \infty} \frac{4x^3 - x + 2}{2x^3 + 7x^2 - 5} \;=\; \frac{4}{2} \;=\; 2. $$If you want to see the mechanics, divide every term by $x^3$:
$$ \frac{4 - \tfrac{1}{x^2} + \tfrac{2}{x^3}}{2 + \tfrac{7}{x} - \tfrac{5}{x^3}} \;\longrightarrow\; \frac{4 - 0 + 0}{2 + 0 - 0} \;=\; 2 \quad \text{as } x \to \infty. $$Example 5 · A standard limit: $\displaystyle\lim_{x \to 0} \frac{\sin(3x)}{x}$
Direct substitution gives $\tfrac{0}{0}$. The expression looks like $\tfrac{\sin x}{x}$ in disguise — but the argument of sine is $3x$, not $x$. Rewrite to expose the pattern by multiplying and dividing by $3$:
$$ \frac{\sin(3x)}{x} \;=\; 3 \cdot \frac{\sin(3x)}{3x}. $$Now let $u = 3x$; as $x \to 0$, $u \to 0$ as well. The remaining fraction approaches $\tfrac{\sin u}{u} \to 1$, so
$$ \lim_{x \to 0} \frac{\sin(3x)}{x} \;=\; 3 \cdot 1 \;=\; 3. $$The general pattern: $\lim_{x \to 0} \tfrac{\sin(kx)}{x} = k$ for any constant $k$. Recognising the standard form saves you from rederiving.