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Limits of Functions (Layered)

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Limits of Functions (Layered)

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18 min read

2026-09-12

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Now · Where this chapter sits

Limits of Functions

One sitting

This chapter is designed to be read start-to-finish in one focused session (roughly 45–60 minutes including the checks). By the end you will be able to read limit notation aloud, evaluate limits with a five-step toolkit, recognize every standard way papers ask about limits, and know exactly what to do next.

Where this chapter sits

The limit is the single idea the rest of calculus is built on: a derivative is a limit of slopes, an integral is a limit of sums, and continuity is a statement about limits. If limits feel solid, the next three chapters get dramatically easier.

You need first:

Sequences and Convergence — the idea of values "settling toward" a target
Functions and Their Types — reading $f(x)$ behavior near a point
Polynomial Operations — factoring, which powers the most common technique

This unlocks:

Continuity — "the limit equals the value"
Derivatives — defined as a limit of difference quotients

How to read the symbols

Every symbol below appears in this chapter. Read this table once now, and come back to it whenever a formula stops making sense — most confusion about limits is actually confusion about notation.

Symbol	Read it aloud as	What it means	Worth remembering
$\lim_{x \to a} f(x)$	"the limit of $f$ of $x$ as $x$ approaches $a$ "	the value $f(x)$ trends toward as $x$ closes in on $a$	lim is from Latin līmes, "boundary" — the value at the edge of the approach
$\to$	"approaches" / "tends to"	gets arbitrarily close to, without needing to arrive	the arrow notation was popularized by G. H. Hardy (1908)
$x \to a^-$	" $x$ approaches $a$ from the left"	only values smaller than $a$ (think: from the minus side)	the sign marks the side, not a negative number
$x \to a^+$	" $x$ approaches $a$ from the right"	only values larger than $a$	—
$\infty$	"infinity"	unbounded growth — a direction, not a number	the lemniscate symbol was introduced by John Wallis (1655)
DNE	"does not exist"	the two sides disagree, oscillate, or blow up	writing " $= \text{DNE}$ " is accepted shorthand in answers
$\tfrac{0}{0}$ , $\tfrac{\infty}{\infty}$ , $1^\infty$ , …	"indeterminate form"	the form alone does not determine the answer — more work needed	indeterminate ≠ undefined: it means "undecided until you look closer"
$\iff$	"if and only if"	each side implies the other	both directions must hold
$f'(x)$	" $f$ prime of $x$ "	the derivative of $f$ (used only in the L'Hôpital section)	Lagrange's prime notation

Layer 1 — Intuition: the trend, not the value

What happens to

f(x) = x^2

x

gets closer and closer to

3

? Watch the numbers approach from both sides:

$x$ (from the left)	$f(x)$	$x$ (from the right)	$f(x)$
$2.9$	$8.41$	$3.1$	$9.61$
$2.99$	$8.9401$	$3.01$	$9.0601$
$2.999$	$8.994$	$3.001$	$9.006$

Both columns are being squeezed toward the same number,

9

. We write

\lim_{x \to 3} x^2 = 9

Here is the idea that makes limits interesting rather than just "plugging in":

The limit ignores what happens AT the point

Take

g(x) = \dfrac{x^2 - 4}{x - 2}

. At

x = 2

this is

\tfrac{0}{0}

— undefined. The function has a hole there. Yet for every

x \neq 2

g(x) = x + 2

, so as

x \to 2

the outputs trend cleanly toward

4

. The limit exists (

= 4

) even though the value

g(2)

does not. A limit describes the approach, never the arrival.

That is why limits can define derivatives: a slope at a single point is a

\tfrac{0}{0}

situation ("rise zero over run zero"), and only the trend of nearby slopes makes sense of it.

Layer 2 — The definition, made precise

Informal definition.

\lim_{x \to a} f(x) = L

means:

f(x)

can be made arbitrarily close to

L

by taking

x

sufficiently close to

a

(from either side), without requiring

x = a

Think of it as a challenge game: a skeptic demands "get

f(x)

within one millionth of

L

", and you must be able to answer "done — for every

x

within this distance of

a

". If you can win the game for every demanded closeness, the limit is

L

. (The formal

\varepsilon

–

\delta

version of this game is beyond what Week 7 requires, but this is exactly what it says.)

One-sided limits

Left-hand limit $\lim_{x \to a^-} f(x)$ : approach using only $x < a$ .
Right-hand limit $\lim_{x \to a^+} f(x)$ : approach using only $x > a$ .

The two-sided limit exists exactly when both sides exist and agree:

\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L

Non-examples: the three ways a limit fails

A concept is only half-learned until you can say what it is not. A limit at

a

fails to exist in exactly three ways:

Failure mode	Example	What goes wrong near the point
Jump — sides disagree	$f(x) = \dfrac{\lvert x \rvert}{x}$ at $x = 0$	left side trends to $-1$ , right side to $+1$
Blow-up — unbounded	$f(x) = \dfrac{1}{x^2}$ at $x = 0$	outputs grow without bound (we may write $\to \infty$ , but no finite limit exists)
Oscillation — never settles	$f(x) = \sin\!\left(\dfrac{1}{x}\right)$ at $x = 0$	the outputs sweep through $[-1, 1]$ forever, faster and faster

Question 1

Type: MCQ | Marks: 1 | Topic: One-sided limits and existence

Let

f(x) = \begin{cases} x + 1 & x < 2 \\ 5 & x = 2 \\ x^2 - 1 & x > 2 \end{cases}

What is

\lim_{x \to 2} f(x)

MCQ

One-sided limits and existence

3 Unit Assessment

3

5

4

The limit does not exist

Layer 3 — The evaluation toolkit

Almost every limit question is solved by walking this ladder in order:

Substitute $x = a$ . Got a plain number? Done — that is the limit.
Got $\tfrac{\text{nonzero}}{0}$ ? The limit is infinite or DNE (check the sign from each side).
Got $\tfrac{0}{0}$ ? Rewrite: factor, rationalize, or match a standard limit — then substitute again.
Limit at $\infty$ of a ratio? Divide by the highest power (or compare degrees).
Still stuck on $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ ? L'Hôpital's rule.

Limit laws

\lim_{x \to a} f(x) = L

and

\lim_{x \to a} g(x) = M

(both finite):

Law	Statement	Caution
Sum	$\lim\,[f(x) + g(x)] = L + M$	—
Difference	$\lim\,[f(x) - g(x)] = L - M$	—
Product	$\lim\,[f(x)\, g(x)] = L\,M$	—
Quotient	$\lim\,\dfrac{f(x)}{g(x)} = \dfrac{L}{M}$	only if $M \neq 0$
Constant multiple	$\lim\, c\,f(x) = c\,L$	—
Power	$\lim\,[f(x)]^n = L^n$	—
Root	$\lim \sqrt[n]{f(x)} = \sqrt[n]{L}$	need $L \ge 0$ for even $n$

These laws are why direct substitution works for polynomials, and for rational, trigonometric, exponential, and logarithmic functions anywhere they are defined.

Standard limits — memorize these five

Standard limit	Note
$\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1$	only in radians
$\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = 0$	companion: $\displaystyle \lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}$
$\displaystyle \lim_{x \to 0} \frac{e^x - 1}{x} = 1$	" $e^x$ looks like $1 + x$ near $0$ "
$\displaystyle \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1$	mirror of the line above
$\displaystyle \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x} = e$	equivalently $\lim_{x \to 0}(1+x)^{1/x} = e$

The scaling move

Papers rarely give you

\frac{\sin x}{x}

raw — they give

\frac{\sin 3x}{x}

. Force the pattern to match by scaling:

\dfrac{\sin 3x}{x} = 3 \cdot \dfrac{\sin 3x}{3x} \to 3 \cdot 1 = 3

. The same move works for

e^{kx}-1

and

\ln(1+kx)

: the answer picks up a factor of

k

Technique anatomy: factoring a 0/0 limit, every step justified

The single most common exam limit is a rational function that gives

\tfrac{0}{0}

. Here is one dissected completely — this is the level of justification you should be able to reproduce:

\lim_{x \to 2} \frac{x^2 - 4}{x - 2}

Step	What we write	Why it is allowed
1. Substitute $x = 2$	$\frac{4 - 4}{2 - 2} = \frac{0}{0}$	always try substitution first; $\frac{0}{0}$ signals "rewrite needed", not "no answer"
2. Factor the numerator	$x^2 - 4 = (x-2)(x+2)$	difference of squares; a $\frac{0}{0}$ from polynomials guarantees $(x - a)$ is a common factor
3. Cancel $(x - 2)$	$\frac{(x-2)(x+2)}{x-2} = x + 2$	valid because the limit only uses $x \neq 2$ , and there division by $(x-2)$ is legal
4. Substitute again	$\lim_{x \to 2} (x+2) = 4$	$x + 2$ is a polynomial — direct substitution applies

Step 3 is the philosophical heart: cancelling is legal precisely because the limit never uses the point itself. The original function and

x+2

differ only at

x = 2

, and the limit cannot see that one point.

Technique: rationalization (the conjugate move)

When a square root causes

\tfrac{0}{0}

, multiply top and bottom by the conjugate (same two terms, opposite sign) to release the hidden factor:

\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} = \lim_{x \to 0} \frac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{x(\sqrt{x+1}+1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+1}+1)} = \lim_{x \to 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{2}

The conjugate works because

(A - B)(A + B) = A^2 - B^2

deletes the square root from the numerator, exposing a factor of

x

to cancel.

Technique: limits at infinity

For

x \to \infty

of a ratio, divide numerator and denominator by the highest power of

x

in the denominator — every

\tfrac{c}{x^k}

term then dies:

\lim_{x \to \infty} \frac{3x^2 + 2x}{4x^2 - 1} = \lim_{x \to \infty} \frac{3 + \frac{2}{x}}{4 - \frac{1}{x^2}} = \frac{3}{4}

The shortcut worth internalizing (and being able to re-derive):

Degrees	$\lim_{x \to \infty} \frac{p(x)}{q(x)}$
$\deg p < \deg q$	$0$
$\deg p = \deg q$	ratio of leading coefficients
$\deg p > \deg q$	$\pm\infty$ (no finite limit)

Technique: L'Hôpital's rule

If substituting gives exactly

\tfrac{0}{0}

\tfrac{\infty}{\infty}

, and

\lim \frac{f'(x)}{g'(x)}

exists, then:

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

Differentiate the numerator and denominator separately — this is not the quotient rule. (The rule is named for Guillaume de l'Hôpital, who published it in 1696; it was actually discovered by his hired tutor Johann Bernoulli — a useful memory hook: someone else does the work, the name on top gets credit, and so does the derivative of the top.)

Example (repeat if still indeterminate):

\lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x} = 0

Each application is legal because the form was

\tfrac{\infty}{\infty}

at that stage. This example also teaches a fact worth keeping: exponentials beat polynomials in any race to infinity.

Example ( $0/0$ , two rounds):

\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}

Technique: the $1^\infty$ form

For

\lim f(x)^{g(x)}

where

f \to 1

and

g \to \infty

, use:

\lim f(x)^{g(x)} = e^{\lim\, [f(x) - 1]\, g(x)}

Example:

\lim_{x \to 0} (1 + 2x)^{1/x} = e^{\lim_{x \to 0} 2x \cdot \frac{1}{x}} = e^2

All indeterminate forms at a glance

Form	Example	Standard treatment
$\tfrac{0}{0}$	$\frac{x^2 - 4}{x - 2}$ at $x = 2$	factor / rationalize / standard limits / L'Hôpital
$\tfrac{\infty}{\infty}$	$\frac{3x^2 + 2x}{4x^2 - 1}$ , $x \to \infty$	divide by highest power / L'Hôpital
$0 \cdot \infty$	$x \ln x$ , $x \to 0^+$	rewrite as $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ , e.g. $\frac{\ln x}{1/x}$
$\infty - \infty$	$\sqrt{x+1} - \sqrt{x}$ , $x \to \infty$	common denominator or conjugate
$1^\infty$	$(1 + \tfrac{1}{x})^x$ , $x \to \infty$	$e^{\lim (f-1)g}$
$0^0$ , $\infty^0$	$x^x$ , $x \to 0^+$	take $\ln$ , resolve, exponentiate back

Question-pattern anatomy

There are only a handful of concepts in this chapter, and papers can only probe them in a handful of shapes. Learn to name the pattern before touching the algebra — recognition is half the marks. These patterns are distilled from the structure of the toolkit above and the archived May 2026 papers (drillable on the checkpoint desk); they are not tied to any live graded work.

#	Pattern	Recognition cue	Attack	The planted trap
1	Disguised substitution	limit looks scary but denominator is nonzero at $a$	just substitute	panicking into L'Hôpital and wrecking a finished answer
2	Factorable $\tfrac{0}{0}$	rational function, both top and bottom vanish at $a$	factor out $(x - a)$ , cancel, substitute	sign slips while factoring; forgetting to re-substitute
3	Conjugate radical	$\sqrt{\cdot} - \text{number}$ over something vanishing	multiply by conjugate	multiplying only the numerator
4	Standard-limit scaling	$\sin kx$ , $e^{kx} - 1$ , or $\ln(1 + kx)$ over $x$	force the exact pattern, collect the factor $k$	$\frac{\sin 5x}{\sin 2x}$ needs the move twice → $\frac{5}{2}$
5	Degree race	rational function, $x \to \infty$	compare degrees / divide by highest power	reading the answer off the constant terms instead of leading terms
6	One-sided / piecewise	piecewise definition, or $\lvert x \rvert$ , near the seam	compute left and right separately, then compare	reporting $f(a)$ instead of the limit; missing that sides disagree
7	$1^\infty$ compound	$(1 + \text{small})^{\text{large}}$	$e^{\lim (f-1) g}$	answering $1$ ("one to any power is one" — false here)
8	L'Hôpital bait	quotient that is not $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$	substitute normally	the question is designed so L'Hôpital gives a clean wrong number

Question 2

Type: MCQ | Marks: 1 | Topic: Standard-limit scaling (pattern 4)

Evaluate

\displaystyle \lim_{x \to 0} \frac{\sin 5x}{\sin 2x}

MCQ

Standard-limit scaling (pattern 4)

3 Unit Assessment

1

\dfrac{5}{2}

\dfrac{2}{5}

The limit does not exist

Question 3

Type: NAT | Marks: 1 | Topic: Factorable 0/0 (pattern 2)

Evaluate

\displaystyle \lim_{x \to 3} \frac{x^2 - 9}{x - 3}

. Enter the value.

MCQ

3 Unit Assessment

Reference-only archive item

The completed export preserved the prompt as an image but not a reusable answer key. The reconstruction below is for study, not scoring.

Study reconstruction

Accepted Answer: 6 Substitution gives

\tfrac{0}{0}

. Factor:

\dfrac{(x-3)(x+3)}{x-3} = x + 3 \to 6

Question 4

Type: MCQ | Marks: 1 | Topic: L'Hôpital bait (pattern 8)

Evaluate

\displaystyle \lim_{x \to 1} \frac{x^2 + 1}{x + 1}

MCQ

L'Hôpital bait (pattern 8)

3 Unit Assessment

1

2

0

The limit does not exist

Common mistakes

Using L'Hôpital on a non-indeterminate form

\lim_{x \to 1} \frac{x^2}{x} = 1

by substitution. L'Hôpital would give

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

— wrong, because the form was

\frac{1}{1}

. The rule's precondition is exactly

\frac{0}{0}

\frac{\infty}{\infty}

Confusing the limit with the value

\lim_{x \to a} f(x)

and

f(a)

are independent pieces of information. Either can exist without the other. (When they exist and agree, that is precisely what the next chapter calls continuity.)

Declaring a two-sided limit from one side

\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)

, the two-sided limit does not exist — full stop. Piecewise and

\lvert x \rvert

questions are built on this.

Using

\frac{\sin x}{x} \to 1

in degrees The standard limit holds only for

x

in radians. In degrees the limit is

\frac{\pi}{180}

— which is why radians are the default in calculus.

L'Hôpital via the quotient rule

The rule says

\frac{f'}{g'}

— top and bottom differentiated separately. Applying the quotient rule to the whole fraction is a different (wrong) object.

Worked exercise set

Twelve exercises spanning every pattern, with full working. Cover the working, attempt each one, then compare — and for each, name the pattern (1–8) before solving. If you want the answers hidden and scored, use the interactive questions above and the drill decks instead.

Q1 (pattern 1): Find $\lim_{x \to 3} (2x + 1)$ .

Direct substitution: $2(3) + 1 = 7$ . $\boxed{7}$

Q2 (pattern 2): Find $\lim_{x \to 2} \frac{x^2 - 3x + 2}{x-2}$ .

Factor: $\frac{(x-1)(x-2)}{x-2} = x-1 \to 1$ . $\boxed{1}$

Q3 (pattern 3): Find $\lim_{x \to 0} \frac{\sqrt{4+x} - 2}{x}$ .

Rationalize: $\frac{(\sqrt{4+x} - 2)(\sqrt{4+x}+2)}{x(\sqrt{4+x}+2)} = \frac{x}{x(\sqrt{4+x}+2)} = \frac{1}{\sqrt{4+x}+2} \to \frac{1}{4}$ . $\boxed{\tfrac{1}{4}}$

Q4 (pattern 5): Find $\lim_{x \to \infty} \frac{5x^2 - 3x + 1}{2x^2 + 7}$ .

Equal degrees → ratio of leading coefficients: $\frac{5}{2}$ . $\boxed{\tfrac{5}{2}}$

Q5 (pattern 4): Find $\lim_{x \to 0} \frac{\sin 3x}{x}$ .

$\frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x} \to 3 \cdot 1 = 3$ . $\boxed{3}$

Q6 (pattern 4): Find $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$ .

L'Hôpital (form $\tfrac{0}{0}$ ): $\lim_{x\to 0} \frac{\sin x}{2x} = \frac{1}{2} \cdot 1 = \frac{1}{2}$ . $\boxed{\tfrac{1}{2}}$

Q7 (pattern 4): Find $\lim_{x \to 0} \frac{e^{2x} - 1}{x}$ .

Scale: $2 \cdot \lim_{x\to 0} \frac{e^{2x} - 1}{2x} = 2 \cdot 1 = 2$ . $\boxed{2}$

Q8 (pattern 4): Evaluate $\lim_{x \to 0} \frac{x^2}{\sin x}$ .

$\frac{x^2}{\sin x} = \frac{x}{\sin x} \cdot x \to 1 \cdot 0 = 0$ . $\boxed{0}$

Q9 (pattern 2): Find $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$ .

Factor: $\frac{(x-1)(x^2+x+1)}{x-1} = x^2 + x + 1 \to 3$ . $\boxed{3}$

Q10 ( $0 \cdot \infty$ ): Find $\lim_{x \to 0^+} x \ln x$ .

Rewrite as $\frac{\ln x}{1/x}$ (form $\frac{-\infty}{\infty}$ ), then L'Hôpital: $\frac{1/x}{-1/x^2} = -x \to 0$ . $\boxed{0}$

Q11 (L'Hôpital): Find $\lim_{x \to 0} \frac{e^x - 1}{\sin x}$ .

Form $\tfrac{0}{0}$ : $\frac{e^x}{\cos x} \to \frac{1}{1} = 1$ . $\boxed{1}$

Q12 (pattern 7): Find $\lim_{x \to 0} (1 + 2x)^{1/x}$ .

$1^\infty$ form: $e^{\lim_{x\to 0} (1 + 2x - 1)\cdot \frac{1}{x}} = e^{\lim 2x/x} = e^2$ . $\boxed{e^2}$

Check yourself — blank-page recall

Close the notes and answer from memory. If one stalls, the section to reread is named — that stall is the point of the exercise, not a failure.

Explain to an imaginary classmate why $\lim_{x \to 2} \frac{x^2-4}{x-2}$ can equal $4$ when the function is undefined at $2$ . (Layer 1)
List the three ways a limit can fail to exist, with one example each. (Layer 2)
Write the five standard limits from memory, and state the one condition attached to the sine limit. (Layer 3)
What exactly must be true before L'Hôpital's rule may be used, and what do you differentiate? (Layer 3)
Name the eight question patterns and the trap planted in each. (Pattern anatomy)

Key formulas — one screen

\boxed{\;\lim_{x\to 0} \frac{\sin x}{x} = 1 \qquad \lim_{x\to 0} \frac{e^x-1}{x} = 1 \qquad \lim_{x\to 0} \frac{\ln(1+x)}{x} = 1 \qquad \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = e\;}

Form met	Move
plain number on substitution	done
$\tfrac{0}{0}$ , polynomial	factor and cancel
$\tfrac{0}{0}$ , radical	conjugate
$\tfrac{0}{0}$ , sin / exp / log	standard limits + scaling
ratio as $x \to \infty$	degree comparison
$\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ , else	L'Hôpital
$1^\infty$	$e^{\lim(f-1)g}$
$0 \cdot \infty$ , $\infty - \infty$	rewrite into a quotient first

Next action

Next chapter: Continuity — limits meeting values.
Drill now: the Maths deck on the Quiz 3 checkpoint desk — attempt first, key after commit.
Spacing (planning suggestion, not a rule): redo the blank-page recall tomorrow (~10 min); redo only your missed exercises in 3 days.

Chapter provenance: authored 2026-09-12 to the layered-chapter standard (docs/roadmaps/2026-09-12-claude-content-ecosystem.md), derived from the 2026-08-01 iitm-study-os import (source URL in front matter). The untouched import is preserved alongside as “Limits of Functions” — imports are source, never destination. Graphs for the Layer 1–2 examples are a known gap — numeric tables stand in until a visual earns its pixels.