Limits

Imagine walking toward a wall, halving the remaining distance with every step. You get closer and closer — 1 metre away, then half a metre, then a quarter — but each step, taken literally, never lands you on the wall. Yet everyone knows where you are heading. A limit is mathematics' precise way of naming that destination: the single value a function approaches as its input approaches some point, even if the function never actually reaches it (or isn't even defined there).

This idea is the bedrock of all of calculus. Derivatives are limits, integrals are limits, and infinite series are limits. Without a solid grasp of limits, the rest of calculus is just formulas to memorize; with it, the whole subject becomes a single coherent story about approaching, approximating, and getting exactly right in the end.

Learning Objectives

By the end of this page, you should be able to:

• Explain in plain language and precise notation what $\lim_{x \to a} f(x) = L$ means.
• Compute left-hand and right-hand (one-sided) limits and use them to decide whether a two-sided limit exists.
• Evaluate limits at infinity and interpret horizontal asymptotes.
• Define continuity at a point using limits and identify where functions break.
• Recognize why $0/0$ is indeterminate and resolve it by factoring, rationalizing, or simplifying.
• Tell the story of why limits had to be invented to make calculus rigorous.

Quick Answer

The limit of $f(x)$ as $x$ approaches $a$ is the value $L$ that $f(x)$ gets arbitrarily close to as $x$ gets arbitrarily close to $a$ (from both sides), without requiring $x = a$. We write $\lim_{x \to a} f(x) = L$. A two-sided limit exists only when the left-hand limit and right-hand limit agree. A function is continuous at $a$ when the limit exists, $f(a)$ is defined, and the two are equal — intuitively, when you can draw it without lifting your pen. Limits let us make sense of expressions like $\frac{0}{0}$, which are indeterminate: the form alone tells you nothing, but simplifying the function first often reveals a perfectly good answer. This "approach without arrival" idea is exactly what powers derivatives and integrals.

Where It Came From

Limits were not invented for their own sake — they were the rescue that saved calculus from a two-thousand-year-old crisis of logic.

The trouble began with Zeno of Elea (around 450 BCE). His paradoxes — Achilles never catching the tortoise, the arrow that is at rest at every instant yet somehow moves — all turned on the same puzzle: how can infinitely many steps, or infinitely small instants, add up to a finite, sensible answer? Greek mathematicians like Archimedes got startlingly close, using a "method of exhaustion" to find areas by squeezing a shape between inscribed and circumscribed polygons. That is a limit in all but name, but the Greeks never made the underlying process rigorous.

The real explosion came in the 1660s–1680s when Isaac Newton and Gottfried Leibniz independently invented calculus. Their results were spectacular — planetary orbits, tangent lines, areas under curves — but their justification was shaky. They computed derivatives by dividing by a quantity they called an infinitesimal: a number treated as nonzero (so you could divide by it) and then, a line later, as zero (so it disappeared). The philosopher George Berkeley skewered this in 1734 in The Analyst, mocking these vanishing quantities as "the ghosts of departed quantities." He was right: the logic was contradictory.

The fix took over a century. Augustin-Louis Cauchy (1820s) recast calculus around the idea of a limit rather than infinitesimals, and Karl Weierstrass (1850s–1870s) finally gave the airtight epsilon-delta definition: $\lim_{x\to a} f(x)=L$ means that for every tolerance $\varepsilon > 0$ you demand around $L$, there is a nearness $\delta > 0$ around $a$ that guarantees it. No ghosts, no dividing by zero — just a precise game of "how close do you need $f(x)$? I can get $x$ close enough." That definition is still the foundation of analysis today.

What a Limit Really Means

Consider $f(x) = \dfrac{x^2 - 1}{x - 1}$. At $x = 1$ this is $\frac{0}{0}$ — undefined. But that says nothing about what happens near $x = 1$. Factor the top:

$f(x) = \frac{(x-1)(x+1)}{x-1} = x + 1 \quad \text{for } x \neq 1.$

For every input except the single forbidden point, $f$ behaves exactly like $x + 1$. So as $x$ approaches 1, $f(x)$ approaches $1 + 1 = 2$:

$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2.$

The graph is the line $y = x + 1$ with a tiny hole punched at $(1, 2)$. The limit fills in the hole — it tells you where the function wants to go, regardless of whether it is allowed to be there. This is the crucial mental shift: the value of the limit and the value of the function at the point are two different questions.

Worked example. Evaluate $\lim_{x \to 3}(2x^2 - 5x + 1)$. Polynomials are continuous everywhere, so we can simply substitute:

$2(3)^2 - 5(3) + 1 = 18 - 15 + 1 = 4.$

Direct substitution works whenever the function is continuous at the point — which is most of the time. It is the broken cases, where substitution gives $\frac{0}{0}$ or a jump, that make limits interesting.

One-Sided Limits

Sometimes a function approaches different values depending on which direction you come from. We write:

• $\lim_{x \to a^-} f(x)$ — the left-hand limit, approaching from values less than $a$.
• $\lim_{x \to a^+} f(x)$ — the right-hand limit, approaching from values greater than $a$.

The two-sided limit $\lim_{x\to a} f(x)$ exists if and only if both one-sided limits exist and are equal.

Worked example. Let $f(x) = \dfrac{|x|}{x}$. For $x > 0$, $|x| = x$, so $f(x) = 1$. For $x < 0$, $|x| = -x$, so $f(x) = -1$. Then:

$\lim_{x \to 0^+} \frac{|x|}{x} = 1, \qquad \lim_{x \to 0^-} \frac{|x|}{x} = -1.$

The one-sided limits disagree, so $\lim_{x\to 0} \frac{|x|}{x}$ does not exist. The graph jumps from $-1$ to $+1$ at the origin — a genuine break, not a fillable hole.

One-sided limits are essential for piecewise functions, and for defining the derivative at endpoints of an interval.

Limits at Infinity and Continuity

Limits at Infinity

Here we ask what happens to $f(x)$ as $x$ grows without bound. This describes the end behavior of a function and reveals horizontal asymptotes.

Worked example. Evaluate $\lim_{x \to \infty} \dfrac{3x^2 + 5x}{2x^2 - 7}$. Divide the top and bottom by the highest power, $x^2$:

$\frac{3x^2 + 5x}{2x^2 - 7} = \frac{3 + \frac{5}{x}}{2 - \frac{7}{x^2}}.$

As $x \to \infty$, the terms $\frac{5}{x}$ and $\frac{7}{x^2}$ shrink to 0:

$\lim_{x \to \infty} \frac{3x^2 + 5x}{2x^2 - 7} = \frac{3 + 0}{2 - 0} = \frac{3}{2}.$

The line $y = \frac{3}{2}$ is a horizontal asymptote. A useful shortcut for rational functions: when the top and bottom have the same degree, the limit at infinity is the ratio of leading coefficients.

Continuity

A function $f$ is continuous at $a$ when all three of these hold:

1. $f(a)$ is defined,
2. $\lim_{x \to a} f(x)$ exists, and
3. $\lim_{x \to a} f(x) = f(a)$.

If any condition fails, there is a discontinuity. A removable discontinuity is a hole (the limit exists but doesn't match $f(a)$, as in our $x+1$ example). A jump discontinuity is where one-sided limits disagree (like $|x|/x$). An infinite discontinuity is a vertical asymptote, as in $f(x) = 1/x$ at $x = 0$.

Worked example. Is $g(x) = \begin{cases} x^2 & x \le 2 \\ 3x - 2 & x > 2 \end{cases}$ continuous at $x = 2$? Check the pieces: left limit is $2^2 = 4$; right limit is $3(2) - 2 = 4$; and $g(2) = 2^2 = 4$. All three agree, so $g$ is continuous at $x = 2$. The pieces meet seamlessly.

Why 0/0 Needs Limits

The expression $\frac{0}{0}$ is called an indeterminate form because its value depends entirely on how the numerator and denominator approach zero. Consider three examples, all of the form $\frac{0}{0}$ at $x=0$:

• $\lim_{x\to 0}\dfrac{x}{x} = 1$
• $\lim_{x\to 0}\dfrac{5x}{x} = 5$
• $\lim_{x\to 0}\dfrac{x^2}{x} = \lim_{x\to 0} x = 0$

Same form, three different answers. That is precisely why you cannot just "plug in" — you must simplify first.

Worked example (rationalizing). Evaluate $\lim_{x\to 0}\dfrac{\sqrt{x+4}-2}{x}$. Substituting gives $\frac{0}{0}$. Multiply by the conjugate:

$\frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} = \frac{(x+4)-4}{x(\sqrt{x+4}+2)} = \frac{x}{x(\sqrt{x+4}+2)} = \frac{1}{\sqrt{x+4}+2}.$

Now substitute safely:

$\lim_{x\to 0}\frac{1}{\sqrt{x+4}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{4}.$

The moral: an indeterminate form is an invitation to do algebra, not a dead end.

Real-World Applications

• Instantaneous rates (physics and engineering): Velocity is the limit of average velocity over shrinking time intervals. Every speedometer, acceleration curve, and rate-of-reaction is a limit made concrete — this is literally the definition of the derivative.
• Areas and totals (integration): The area under a curve is the limit of the total area of ever-thinner rectangles. Fuel consumed, charge accumulated, and probability under a bell curve are all computed this way.
• Economics: Marginal cost and marginal revenue are limits of cost/revenue change as production increases by an ever-smaller amount, guiding pricing and output decisions.
• Medicine and pharmacology: Drug concentration models use limits at infinity to find the steady-state level a repeated dose approaches over long-term treatment.
• Engineering stability: Whether a control system settles or blows up is a question about the limit of its response as time goes to infinity.
• Computer graphics and numerical methods: Iterative algorithms (Newton's method, rendering refinement) are designed so their output converges — approaches a limit — to the true answer.

Common Mistakes

Mistake 1: Thinking the limit is just $f(a)$. Many students believe $\lim_{x\to a} f(x)$ always equals $f(a)$. It is often true (that's continuity), but the whole point of limits is the cases where it fails — holes, jumps, and asymptotes. The limit asks where the function is heading near $a$, not what it is at $a$. Always check whether the function is actually continuous before substituting.

Mistake 2: Believing $\frac{0}{0}$ equals 0, or 1, or "undefined, so no limit." As shown above, $\frac{0}{0}$ can approach any value. It is indeterminate, meaning the form alone is inconclusive. The correction: simplify the expression (factor, rationalize, cancel) until you can substitute cleanly.

Mistake 3: Assuming every two-sided limit exists. If the left and right limits differ, the two-sided limit does not exist — you cannot average them or pick one. Correction: for piecewise functions and anything with absolute values, always compute both one-sided limits separately before concluding.

Mistake 4: Writing "$\lim = \infty$" and calling it a valid limit. Saying a limit "equals infinity" is shorthand for the function grows without bound — the limit does not exist as a finite number. Be clear about which you mean; on many exams "$\infty$" and "does not exist" must be distinguished carefully.

Comparison and Connections

Concept	What it asks	Key requirement
Limit at $a$	Where does $f$ head near $a$?	Left and right limits agree
Value $f(a)$	What is $f$ at $a$?	$a$ in the domain
Continuity at $a$	Do the two above match?	Limit exists and equals $f(a)$
One-sided limit	Heading from one side only	Only that side's behavior
Limit at infinity	End behavior of $f$	Behavior as $x$ grows unboundedly

The deepest connection is forward: the derivative is defined as the limit $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, which is a $\frac{0}{0}$ form resolved by exactly the algebra you practiced here. The definite integral is a limit of Riemann sums. Limits are not one topic among many in calculus — they are the language the entire subject is written in.

Practice Questions

Recall

State the three conditions for $f$ to be continuous at $x = a$.

Answer: (1) $f(a)$ is defined; (2) $\lim_{x\to a} f(x)$ exists; (3) $\lim_{x\to a} f(x) = f(a)$.

Understanding

Why does the existence and value of $\lim_{x\to 2}f(x)$ not depend on the value of $f(2)$?

Answer: The limit is defined only in terms of values of $f(x)$ for $x$ near 2 but not equal to 2. The single point $x=2$ is explicitly excluded, so $f(2)$ can be anything — undefined, or a stray value — without changing where the function approaches.

Application

Evaluate $\lim_{x\to 2}\dfrac{x^2 - 4}{x - 2}$.

Answer: Factor: $\frac{(x-2)(x+2)}{x-2} = x+2$ for $x\neq 2$. So the limit is $2 + 2 = 4$.

Analysis

For $f(x) = \begin{cases} x + 1 & x < 1 \\ 4 & x = 1 \\ 2x & x > 1 \end{cases}$, determine $\lim_{x\to 1}f(x)$ and whether $f$ is continuous at $x=1$.

Answer: Left limit: $1 + 1 = 2$. Right limit: $2(1) = 2$. They agree, so $\lim_{x\to 1}f(x) = 2$. But $f(1) = 4 \neq 2$, so $f$ is not continuous at $x=1$ — it has a removable discontinuity (the point sits above the hole).

FAQ

Does the function have to be defined at $a$ for the limit to exist? No. In fact the most important limits — like the derivative — are at points where the function is $\frac{0}{0}$. The limit only cares about the neighborhood around $a$, not $a$ itself.

What's the difference between "the limit is infinite" and "the limit does not exist"? Both mean there is no finite limiting value. "$\lim = \infty$" additionally tells you how it fails: the function grows without bound in a consistent direction. If a function oscillates or jumps instead, we just say the limit does not exist.

Can a limit exist even if the left and right sides give different formulas? Yes — what matters is the values, not the formulas. If a piecewise function's two pieces approach the same number at the boundary (as in the $x^2$ vs $3x-2$ example), the limit exists even though the formulas differ.

Why can't I just plug in the number every time? You can whenever the function is continuous there — which is most polynomials, roots, exponentials, and trig functions on their domains. Plugging in fails only at discontinuities and indeterminate forms, and those are exactly the cases worth studying.

What is L'Hôpital's rule, and do I need it here? L'Hôpital's rule is a shortcut for $\frac{0}{0}$ and $\frac{\infty}{\infty}$ limits using derivatives. It's powerful but comes later — everything on this page can be done with algebra (factoring, canceling, rationalizing), which builds the understanding L'Hôpital assumes.

How does the epsilon-delta definition connect to "getting close"? $\varepsilon$ is the tolerance you demand around the answer $L$; $\delta$ is how close to $a$ you must keep $x$ to stay within that tolerance. Saying the limit is $L$ means: no matter how tight a tolerance $\varepsilon$ you pick, some $\delta$ works. It is "getting close" made into an unbreakable guarantee.

Quick Revision

• $\lim_{x\to a} f(x) = L$: $f(x)$ approaches $L$ as $x$ approaches $a$, ignoring $x = a$ itself.
• Two-sided limit exists $\iff$ left limit $=$ right limit.
• Direct substitution works when $f$ is continuous at $a$.
• $\frac{0}{0}$ is indeterminate — factor, cancel, or rationalize, then substitute.
• Rational functions at infinity: same top/bottom degree $\Rightarrow$ ratio of leading coefficients; smaller top degree $\Rightarrow$ 0.
• Continuity at $a$: $f(a)$ defined, limit exists, and they are equal.
• Discontinuity types: removable (hole), jump (sides differ), infinite (asymptote).
• History: Zeno's paradoxes $\to$ Newton/Leibniz's infinitesimals $\to$ Berkeley's critique $\to$ Cauchy/Weierstrass's epsilon-delta rigor.

Related Topics

Prerequisites

• Functions
• Quadratic Equations

Related Topics

• Calculus Overview
• Graphs and Periodic Motion

Next Topics

• Derivatives
• Integrals
• Applications of Calculus