2.2 The Limit of a Function

Big idea

The previous section used limits informally to define a tangent slope and an area. This section makes the object itself the subject: what a limit is, when one exists, and what the notation commits you to.

The central discipline is a habit of reading. The statement $\lim_{x \to a} f(x) = L$ is a claim about the values of $f$ at inputs close to $a$ but different from $a$. It says nothing about $f(a)$. The function may be undefined there, or defined with a value nowhere near $L$, and neither affects the limit. That separation looks pedantic until you remember where limits came from: every difference quotient in calculus is undefined at exactly the point you care about, so a notion of value that depended on the point would be useless.

Existence is the other half of the story. A limit exists when the values approach one number, and the approach has to work from both directions. Splitting the approach into its two halves gives the one-sided limits, and the rule connecting them is simple: the two-sided limit exists precisely when both one-sided limits exist and agree. Most of the interesting failures come from this rule. A function can jump, so the two sides settle on different numbers. It can blow up, so neither side settles anywhere. It can oscillate forever, so the values never commit at all.

Blowing up gets its own notation. Writing $\lim_{x \to a} f(x) = \infty$ does not name a number the function reaches; it reports a specific kind of nonexistence, namely that the values grow past every bound. The payoff is that this report identifies vertical asymptotes exactly, which turns a graphing habit into a computation you can justify.

Decoder

The limit of a function at a point is $L$ when every input close enough to that point, but not equal to it, produces an output close to $L$, and the two-sided limit exists exactly when the limits from the left and from the right both exist and are equal.

The phrase doing the work is “but not equal to it.” The point is deleted from the conversation on purpose. Any argument you build about a limit must survive having $f(a)$ redefined arbitrarily, so any step that uses $f(a)$ is invalid.

“Close enough” is the other phrase, and it is deliberately vague here because making it precise is the job of a later section. For now, read it as a promise that can be kept at any accuracy: name a tolerance around $L$, and there is a window around $a$ inside which the function honors it.

The classic mistake is to evaluate and stop. Substituting $x = a$ is a legitimate shortcut only for functions that are continuous there, a condition you have not yet earned and which fails for exactly the quotients calculus cares about. When substitution produces $0/0$, that is not an answer; it is a signal that the function needs rewriting before the limit becomes visible.

Definitions and results

Limit, informally. $\lim_{x \to a} f(x) = L$ means the values $f(x)$ can be made as close to $L$ as you like by taking $x$ close enough to $a$ on either side, with $x \ne a$ throughout. If no single $L$ works, the limit does not exist.

Irrelevance of the point. The limit at $a$ does not depend on whether $f(a)$ is defined or on what it equals. Two functions that agree at every input except $a$ have the same limit at $a$.

Left-hand limit. $\lim_{x \to a^-} f(x) = L$ means the values approach $L$ as $x$ approaches $a$ through inputs smaller than $a$. The right-hand limit $\lim_{x \to a^+} f(x)$ is the same statement for inputs larger than $a$.

The two-sided test. $\lim_{x \to a} f(x) = L$ holds if and only if $\lim_{x \to a^-} f(x) = L$ and $\lim_{x \to a^+} f(x) = L$. One side agreeing is never enough, and this is the standard way to show a limit fails.

Infinite limits. $\lim_{x \to a} f(x) = \infty$ means the values grow beyond every bound as $x$ approaches $a$ from either side; $-\infty$ means they fall below every bound. The limit does not exist in these cases. The notation records how it fails.

Vertical asymptote. The line $x = a$ is a vertical asymptote of $f$ when at least one of the one-sided limits at $a$ is $\infty$ or $-\infty$. For a rational function in lowest terms, this happens at every zero of the denominator.

Reading the sign of a blow-up. Near a simple zero of the denominator, the numerator has a fixed nonzero sign while the denominator changes sign as you cross the point, so the two one-sided limits are $\infty$ on one side and $-\infty$ on the other. When the denominator does not change sign, as with a squared factor, both sides run the same way.

Failure by oscillation. A function can fail to have a limit without jumping or blowing up. The values of $\sin(1/x)$ run through the whole interval from $-1$ to $1$ infinitely often in every window around $0$, so they never settle, and $\lim_{x \to 0} \sin(1/x)$ does not exist.

Worked examples

A hole in the graph

Let $f(x) = \dfrac{x^2 - 25}{x - 5}$. At $x = 5$ both parts are zero, so $f(5)$ is undefined. For every other input, factoring gives

$$ f(x) = \frac{(x-5)(x+5)}{x-5} = x + 5 \qquad (x \ne 5) $$

The rewritten form is legal precisely because the limit ignores $x = 5$, which is the only input where the two expressions disagree. The values run $9.9$ at $x = 4.9$ and $10.1$ at $x = 5.1$, so

$$ \lim_{x \to 5} f(x) = 10 $$

The graph is the line $y = x + 5$ with a single point punched out at height $10$. The limit sees the line; it never looks at the puncture.

When the value and the limit disagree

Define

$$ g(x) = \begin{cases} x + 1, & x \ne 2 \\ 7, & x = 2 \end{cases} $$

Approaching $2$ from either side, the relevant inputs are all in the first branch, so the outputs approach $3$. The limit is $3$ while $g(2) = 7$. Nothing is contradictory: one number describes the neighborhood and the other describes a single isolated instruction. This function is the cleanest counterexample to the habit of computing limits by substitution.

A jump, seen through one-sided limits

Define

$$ h(x) = \begin{cases} 2x + 1, & x < 1 \\ 5 - x, & x \ge 1 \end{cases} $$

From the left, only the first branch is in play, and $2x + 1$ approaches $3$. From the right, only the second branch is in play, and $5 - x$ approaches $4$. Both one-sided limits exist:

$$ \lim_{x \to 1^-} h(x) = 3, \qquad \lim_{x \to 1^+} h(x) = 4 $$

They differ, so $\lim_{x \to 1} h(x)$ does not exist. The function value $h(1) = 4$ is not a tiebreaker; it plays no part in the test.

A vertical asymptote and its two sides

Let $k(x) = \dfrac{x + 1}{x - 3}$. The denominator vanishes at $x = 3$ and the numerator does not, so the factor cannot cancel and the line $x = 3$ is a vertical asymptote.

Approaching from the right, the numerator is near $4$ and the denominator is a small positive number, so the quotient is large and positive. From the left, the denominator is a small negative number, so the quotient is large and negative:

$$ \lim_{x \to 3^+} k(x) = \infty, \qquad \lim_{x \to 3^-} k(x) = -\infty $$

Check with numbers. At $x = 3.01$ the value is $4.01 / 0.01 = 401$, and at $x = 2.99$ it is $3.99 / (-0.01) = -399$. Both match the signs predicted.

Contrast this with $\dfrac{x+1}{(x-3)^2}$. The squared denominator stays positive on both sides, so the values run to $\infty$ from both directions and the two-sided limit is $\infty$.

Practice

Begin with graphs and tables. You are shown a curve or a list of values and asked what the outputs approach, including cases where the function has a hole or a value that disagrees with the limit.

Then separate the two directions. These drills ask for left-hand and right-hand limits of piecewise functions and for the verdict on whether the two-sided limit exists.

Last, the unbounded cases. You locate vertical asymptotes and decide which way the function runs on each side of them.

Quiz

Six items covering limits read from graphs, one-sided limits of piecewise functions, and infinite limits at vertical asymptotes.