Contents / Precalculus / Introduction to Limits

Chapter 9

Introduction to Limits

Limit notation and intuition as a bridge from precalculus into calculus.

Introduction

Limit notation and intuition as a bridge from precalculus into calculus.

9.1Limit (mathematics)

The limit is the central idea that makes calculus work: the value the function approaches as input approaches a point (even if the function is not defined or different there).

Limits (Introduction)

Before any formulas, here is the picture. Imagine walking along the graph of a function toward a particular $x$ -value. The limit is the height you are heading toward — the value the graph seems to be aiming at — regardless of whether the graph actually arrives at that height when you land on the point. There might be a hole there, or the point might be lifted somewhere else; the limit ignores that and reports the trend.

Definition 9.1 (Limit of a function). We write

x \to a lim f (x) = L

if $f (x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to $a$ (with $x \neq = a$ ). Read it as: "the limit of $f (x)$ , as $x$ approaches $a$ , equals $L$ . "

The phrase " $x \neq = a$ " is essential — the limit is entirely about the neighborhood around $a$ , never about the single point $a$ itself. This is exactly why a limit can exist at a point where the function is undefined.

We can also approach $a$ from one side only.

Definition 9.2 (One-sided limits). The left-hand limit $lim_{x \to a^{-}} f (x)$ looks only at inputs slightly less than $a$ ; the right-hand limit $lim_{x \to a^{+}} f (x)$ looks only at inputs slightly greater than $a$ .

Theorem 9.3 (Two-sided limits from one-sided limits). The two-sided limit exists if and only if both one-sided limits exist and are equal:

x \to a^{-} lim f (x) = x \to a^{+} lim f (x) = L ⟺ x \to a lim f (x) = L .

If the two sides disagree (a jump), the two-sided limit does not exist. This is the standard way of showing that a limit fails to exist.

Infinite limits and limits at infinity are two different ideas that share the $\infty$ symbol. An infinite limit like $lim_{x \to 0} \frac{1}{x ^{2}} = \infty$ means the output grows without bound as $x$ nears a finite point. A limit at infinity like $lim_{x \to \infty} \frac{1}{x} = 0$ describes the end behavior as $x$ itself runs off to $\pm \infty$ . They call for different techniques, so keep them apart.

Why limits matter. Once you have limits you can define the rest of calculus rigorously:

Continuity at $a$ means $lim_{x \to a} f (x) = f (a)$ — the trend, the destination, and the actual value all agree. This requires three things at once: $f (a)$ is defined, the limit exists, and the two are equal.
Derivatives are limits of average rates of change, $f^{'} (a) = lim_{h \to 0} \frac{f ( a + h ) - f ( a )}{h}$ , giving instantaneous rate.
Integrals are limits of sums of thin rectangles, giving accumulated area.

How limits are evaluated. The techniques come in a definite order of preference.

Method 9.4 (Evaluating a limit).

Direct substitution — if $f$ is continuous at $a$ , just plug in: $lim_{x \to a} f (x) = f (a)$ .
Factor and cancel — when substitution gives the indeterminate form $\frac{0}{0}$ , factor and remove the common factor causing it.
Rationalize — multiply by a conjugate to clear a $\frac{0}{0}$ involving square roots.
Known special limits — most famously $lim_{x \to 0} \frac{s i n x}{x} = 1$ (with $x$ in radians), the engine behind the derivatives of the trig functions.

A form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ is called indeterminate: it does not mean the limit fails to exist, only that this particular substitution didn't settle the question and more work is needed.

Intuition. A limit is like reading a car's GPS as it coasts toward a stop sign. The GPS shows where the car is heading — the corner — even if the car never quite touches the line, or gets teleported to a different spot at the last instant. You care about the destination the motion points to, not the single frozen snapshot at the sign. A one-sided limit is just checking where the car is heading when it's rolling in from the left versus rolling in from the right; if those two answers disagree, there is no single spot it is aiming for.

Example 9.5 (Factor and cancel). Evaluate $x \to 2 lim \frac{x ^{2} - 4}{x - 2}$ .

Solution. Try direct substitution: numerator $= 2^{2} - 4 = 0$ , denominator $= 2 - 2 = 0$ . This gives $\frac{0}{0}$ , an indeterminate form — so substitution alone can't decide it. Factor the numerator using the difference of squares: $x^{2} - 4 = (x - 2) (x + 2)$ . For every $x \neq = 2$ we may cancel the common factor $(x - 2)$ : $\frac{( x - 2 ) ( x + 2 )}{x - 2} = x + 2$ . (We only need $x$ near $2$ , not equal to $2$ , so this cancellation is legal.) Now the limit is of a continuous function, so substitute: $lim_{x \to 2} (x + 2) = 2 + 2 = 4$ .

Sanity check: at $x = 1.99$ , $\frac{x ^{2} - 4}{x - 2} = \frac{3.9601 - 4}{- 0.01} = \frac{- 0.0399}{- 0.01} = 3.99$ , and at $x = 2.01$ it is $\frac{4.0401 - 4}{0.01} = 4.01$ — both hugging $4$ from either side. So $lim_{x \to 2} \frac{x ^{2} - 4}{x - 2} = 4$ , even though $f (2)$ itself is undefined (a hole in the graph).□

Example 9.6 (Rationalizing a root). Evaluate $x \to 0 lim \frac{x + 9 - 3}{x}$ .

Solution. Substitution gives $\frac{9 - 3}{0} = \frac{0}{0}$ , indeterminate. Multiply top and bottom by the conjugate $x + 9 + 3$ (this equals $1$ , so it changes the form but not the value):

\frac{x + 9 - 3}{x} \cdot \frac{x + 9 + 3}{x + 9 + 3} = \frac{( x + 9 ) - 9}{x ( x + 9 + 3 )} .

The numerator simplifies to

x

, and for

x \neq = 0

we cancel it:

\frac{x}{x ( x + 9 + 3 )} = \frac{1}{x + 9 + 3}

. This is continuous at

0

, so substitute:

\frac{1}{9 + 3} = \frac{1}{6}

Sanity check: at $x = 0.01$ , $\frac{9.01 - 3}{0.01} \approx \frac{3.001666 - 3}{0.01} \approx 0.1667 \approx \frac{1}{6}$ . ✓□

Example 9.7 (A jump: when the two sides disagree). Does $x \to 1 lim f (x)$ exist for $f (x) = {x + 1 x^{2} + 3 x < 1 x \geq 1$ ?

Solution. Approach from the left ( $x < 1$ ), where $f (x) = x + 1$ : $lim_{x \to 1^{-}} f (x) = 1 + 1 = 2$ . Approach from the right ( $x \geq 1$ ), where $f (x) = x^{2} + 3$ : $lim_{x \to 1^{+}} f (x) = 1^{2} + 3 = 4$ . The one-sided limits disagree ( $2 \neq = 4$ ), so there is no single value the function heads toward.

Conclusion: $lim_{x \to 1} f (x)$ does not exist — the graph has a jump of height $4 - 2 = 2$ at $x = 1$ . (Note $f (1) = 4$ is defined; a value existing at the point does not rescue a two-sided limit when the sides disagree.)□

Example 9.8 (A continuity check). Is $g (x) = \frac{x ^{2} - 9}{x - 3}$ continuous at $x = 3$ ?

Solution. Check $g (3)$ : it is $\frac{0}{0}$ , so $g$ is undefined at $3$ . Continuity fails immediately — one of the three requirements (a defined value) is missing. The limit itself still exists: $\frac{x ^{2} - 9}{x - 3} = \frac{( x - 3 ) ( x + 3 )}{x - 3} = x + 3$ for $x \neq = 3$ , so $lim_{x \to 3} g (x) = 6$ . Because the limit exists ( $6$ ) but $g (3)$ does not, this is a removable discontinuity: defining $g (3) = 6$ would plug the hole and make $g$ continuous there.

Sanity check: the trend value $6$ matches $x + 3$ at $x = 3$ , and $g$ is undefined only at the single removable point. ✓□

Example 9.9 (A limit at infinity). Evaluate $x \to \infty lim \frac{3 x ^{2} + 2 x}{x ^{2} - 1}$ .

Solution. This is a limit at infinity, so use the "divide by the highest power" technique rather than substitution. The highest power in the denominator is $x^{2}$ ; divide every term by $x^{2}$ :

\frac{3 x ^{2} + 2 x}{x ^{2} - 1} = \frac{3 + \frac{2}{x}}{1 - \frac{1}{x ^{2}}} .

x \to \infty

, the shrinking terms

\frac{2}{x} \to 0

and

\frac{1}{x ^{2}} \to 0

. So the expression approaches

\frac{3 + 0}{1 - 0} = 3

Sanity check: at $x = 1000$ , $\frac{3 000 000 + 2000}{1 000 000 - 1} \approx \frac{3 002 000}{999 999} \approx 3.002$ , closing in on $3$ . This also matches the rule of thumb for rational functions with equal top/bottom degrees: the limit is the ratio of leading coefficients, $\frac{3}{1} = 3$ . ✓□

Example 9.10 (The special trigonometric limit). Evaluate $x \to 0 lim \frac{sin 3 x}{x}$ .

Solution. Substitution gives $\frac{s i n 0}{0} = \frac{0}{0}$ , indeterminate. We want the argument of $sin$ to match the denominator so we can use $lim_{u \to 0} \frac{s i n u}{u} = 1$ . Multiply and divide by $3$ : $\frac{sin 3 x}{x} = 3 \cdot \frac{sin 3 x}{3 x}$ . Let $u = 3 x$ ; as $x \to 0$ , $u \to 0$ too, so $\frac{sin 3 x}{3 x} \to 1$ . Therefore $lim_{x \to 0} \frac{sin 3 x}{x} = 3 \cdot 1 = 3$ .

Sanity check (radians): at $x = 0.001$ , $\frac{s i n ( 0.003 )}{0.001} \approx \frac{0.0029999955}{0.001} \approx 3.0000$ . ✓□

Common Mistakes

Limit vs function value: The limit can exist even when $f (a)$ is undefined or different (removable discontinuity). Always distinguish $lim_{x \to a} f (x)$ from $f (a)$ .
One-sided limits: For limits at endpoints or jumps, the left and right limits must agree; otherwise the two-sided limit does not exist — even if $f (a)$ happens to be defined.
Indeterminate forms: $\frac{0}{0}$ or $\frac{\infty}{\infty}$ require algebraic manipulation or other techniques (L'Hôpital later); do not conclude "does not exist" immediately.
Forgetting to check one-sided at discontinuities: Jump discontinuities have different left/right limits, so always check both sides there.
Plugging in when there is a hole: If substitution gives $\frac{0}{0}$ , simplify (factor, cancel, or rationalize) first — the raw substitution is not the answer.
Limits at infinity vs at a point: Different techniques — divide by the highest power of $x$ for limits at $\infty$ , but substitute or factor for limits at a finite point.
Degrees instead of radians in $\frac{s i n x}{x}$ : The special limit $lim_{x \to 0} \frac{s i n x}{x} = 1$ only holds when $x$ is measured in radians.

Exercises for §9.1 →

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

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