The Limit of a Function

Two geometric problems motivate the entire subject, and both run into the same obstacle: they ask for the value of a ratio at a point where the ratio is undefined. The resolution is the limit, a value that a quantity approaches as its input approaches a target, without the input ever reaching that target.

Two problems that need a limit

The tangent problem

The slope of a straight line is $Δ y /Δ x$ , computed from any two points on it. A curve has no single slope, but it has a tangent line at each point, and we want that tangent's slope. The difficulty is that a tangent touches the curve at only one point, and one point does not determine a slope.

The workaround uses a nearby second point. Fix $P = (1, 1)$ on the parabola $y = x^{2}$ and let $Q = (x, x^{2})$ be another point on it. The line $P Q$ is a secant, and its slope is a genuine two-point quotient,

m_{P Q} = \frac{x ^{2} - 1}{x - 1} = x + 1 (x \neq = 1) .

As $Q$ slides toward $P$ along the curve, the secant pivots toward the tangent, and its slope approaches the tangent's slope.

Secant lines through

P

and a moving point

Q

pivot toward the tangent as

Q

approaches

P

; their slopes approach the tangent slope.

Substituting values of $x$ approaching $1$ gives $m_{P Q}$ approaching $2$ ; the tangent slope is $2$ , even though $m_{P Q}$ is undefined at $x = 1$ itself.

Worked example (secant slopes approaching the tangent). Compute $m_{P Q} = x + 1$ for $Q$ sliding toward $P = (1, 1)$ on $y = x^{2}$ .

$x$	$2$	$1.5$	$1.1$	$1.01$	$1.001$
$m_{P Q} = x + 1$	$3$	$2.5$	$2.1$	$2.01$	$2.001$

Values from the left ( $x = 0.999, 0.99, \dots$ ) approach $2$ the same way. The secant slopes close in on $2$ from both sides, so the tangent at $P$ has slope $2$ .

The velocity problem

The same structure governs instantaneous velocity. A body falling from rest travels $s (t) = 4.9 t^{2}$ meters in $t$ seconds. Its average velocity over the interval from $t = 5$ to $t = 5 + h$ is a difference quotient,

\frac{s ( 5 + h ) - s ( 5 )}{h} = \frac{4.9 ( 5 + h ) ^{2} - 4.9 ( 5 ) ^{2}}{h} = 49 + 4.9 h .

At $h = 0$ this is undefined, but as $h$ shrinks the average velocities approach $49$ , and $49$ m/s is the instantaneous velocity at $t = 5$ .

Worked example (average velocity approaching instantaneous). Tabulate the average velocity $49 + 4.9 h$ over $[5, 5 + h]$ for shrinking $h$ .

$h$ (s)	$1$	$0.1$	$0.01$	$0.001$
avg. velocity (m/s)	$53.9$	$49.49$	$49.049$	$49.0049$

The averages descend toward $49$ as $h \to 0^{+}$ , and the same values arise for $h < 0$ ; the instantaneous velocity at $t = 5$ is $49$ m/s.

Tangent slope and instantaneous velocity are the same operation on different quantities: a difference quotient evaluated in the limit as the increment vanishes.

The intuitive definition

The phrase but not equal to $a$ is essential: the limit describes the approach to $a$ and is indifferent to the value $f (a)$ , which may equal $L$ , differ from $L$ , or fail to exist. All three cases carry the same limit.

In all three cases the limit at

a

equals

L

f (a)

may match

L

(left), differ from it (middle), or be undefined (right).

Consider

g (x) = \frac{x ^{2} - 1}{x - 1},

undefined at $x = 1$ . For every other $x$ it equals $x + 1$ , so a table of values converges on $2$ from both sides.

$x \to 1^{-}$	$g (x)$	$x \to 1^{+}$	$g (x)$
$0.9$	$1.9$	$1.1$	$2.1$
$0.99$	$1.99$	$1.01$	$2.01$
$0.999$	$1.999$	$1.001$	$2.001$

A table converging on the limit from both sides, and the graph with an open circle marking that

g

is undefined at

x = 1

though the limit is

2

Tables suggest a limit but do not prove one; a function can mislead a coarse table by oscillating between the sampled points. The precise definition is what settles a limit for certain, with tables reduced to forming the guess a proof then confirms.

A table used with care still gives the right guess when the function is well behaved between samples.

Worked example (guessing a limit from a table). Estimate $x \to 0 lim \frac{sin x}{x}$ , with $x$ in radians.

The quotient is undefined at $0$ ( $0/0$ ). Tabulating values symmetric about $0$ :

$x$	$\pm 0.5$	$\pm 0.1$	$\pm 0.05$	$\pm 0.01$
$sin (x) / x$	$0.958851$	$0.998334$	$0.999583$	$0.999983$

The values climb toward $1$ from both sides, so the guess is $lim_{x \to 0} (sin x) / x = 1$ . A geometric squeeze argument proves it exactly in the differentiation rules, where it delivers the derivative of $sin x$ .

The oscillation trap is not hypothetical. Sampling $sin (π / x)$ at the reciprocals of integers reads $0$ every time, which invites the wrong guess.

y = sin (π / x)

oscillates between

- 1

and

1

ever faster as

x \to 0

; the values never settle, so the limit at

0

does not exist.

One-sided limits

Approaching $a$ from the left and from the right can give different results, so it helps to name each side.

The two-sided limit exists exactly when the one-sided limits agree.

When the one-sided limits differ, the two-sided limit does not exist. The Heaviside function

H (t) = {01 if t < 0, if t \geq 0,

models a switch thrown at $t = 0$ . Its left limit is $0$ and its right limit is $1$ , so $lim_{t \to 0} H (t)$ does not exist: the graph jumps.

A jump discontinuity: the left limit is

0

and the right limit is

1

, so the two-sided limit at

0

does not exist.

A piecewise function is tested for a limit at each junction by comparing the one-sided limits computed from the formulas on either side.

Infinite limits

A limit can fail because the function grows without bound rather than because two sides disagree. For $f (x) = 1/ x^{2}$ , values near $x = 0$ become arbitrarily large and positive.

The symbol $\infty$ is not a number, and the statement $lim_{x \to a} f (x) = \infty$ does not assert that a limit exists in the ordinary sense; it is shorthand for a specific mode of non-existence, unbounded growth. One-sided infinite limits are defined the same way. The behavior of $1/ x$ near $0$ splits by side:

x \to 0^{+} lim \frac{1}{x} = \infty, x \to 0^{-} lim \frac{1}{x} = - \infty.

An infinite limit locates a vertical feature of the graph.

Near a vertical asymptote

x = a

the graph climbs without bound; here the two-sided limit is

+ \infty

The sign of an infinite limit is read off the sign of numerator and denominator as the input approaches from one side.

Rational functions produce vertical asymptotes at the zeros of the denominator that do not cancel, and $tan x = sin x / cos x$ has one at every odd multiple of $π /2$ , where $cos x = 0$ . Locating these asymptotes is a standard step in curve sketching.

Stewart, §1.5 — the intuitive definition of a limit, the role of one-sided limits, and the interpretation of infinite limits and vertical asymptotes. ↩