Day 1: Limits & Algebraic Limit Laws

Estimating Limits: Notation, Graphs, and Tables

Before doing any algebra, you need to be able to read a limit off a graph or a table of values — this intuitive picture is the foundation everything else in Calculus is built on.

Core Theorem

A limit describes the

y

-value

f(x)

approaches as

x

approaches

c

— not necessarily the value AT

c

.

\lim_{x \to c^-} f(x)

is the value

f(x)

approaches from the LEFT.

\lim_{x \to c^+} f(x)

is the value

f(x)

approaches from the RIGHT.
The two-sided limit

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

exists if and only if

\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Use the table below to estimate

\lim_{x \to 1} f(x)

, where

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

.

x	f(x)
0.9	1.9
0.99	1.99
0.999	1.999
1	undefined
1.001	2.001
1.01	2.01
1.1	2.1

How to think about it

The limit only cares about the trend as

x

approaches

1

— not the value at

x = 1

, which the "undefined" row is there to distract you. Cover that row, then check whether the rows just below and just above it are heading for the same number.

Key idea: Look at what $f(x)$ is approaching as $x$ gets closer to 1 from both sides — ignore the fact that $x=1$ itself is undefined.

Full solution

1. Step 1: Read the Left Side

As

x \to 1^-

(0.9, 0.99, 0.999),

f(x)

is approaching 2.

2. Step 2: Read the Right Side

As

x \to 1^+

(1.1, 1.01, 1.001),

f(x)

is also approaching 2.

3. Step 3: Conclude

Since both sides approach the same value,

\lim_{x \to 1} f(x) = 2

, even though

f(1)

itself is undefined.

Worked Example 02Medium

Determine if

\lim_{x \to 3} \frac{|x-3|}{x-3}

exists.

How to think about it

Absolute value means the expression is really two different formulas —

|x-3| = x-3

for

x > 3

, and

-(x-3)

for

x < 3

. So don't plug in: compute each one-sided limit separately. The two-sided limit exists only if they match — and here they won't.

Key idea: Absolute value functions are 'V-shaped' and often have different behavior on either side of the vertex — check the left and right sides separately.

Full solution

1. Step 1: Test the Right Side

For

x > 3

,

|x-3| = x-3

. Therefore,

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

.

2. Step 2: Test the Left Side

For

x < 3

,

|x-3| = -(x-3)

. Therefore,

\lim_{x \to 3^-} \frac{-(x-3)}{x-3} = -1

.

3. Step 3: Compare and Conclude

Since the left limit (

-1

) does not equal the right limit (

1

), the two-sided limit does not exist (DNE).

Step-by-Step SOP

1
Check Left and Right Separately
For piecewise, absolute-value, or table-based problems, always evaluate $\lim_{x \to c^-}$ and $\lim_{x \to c^+}$ independently first.
2
Compare, Then Conclude
If the two one-sided limits match, that shared value is the limit. If they don't match, the limit does not exist (DNE).

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Compare the behavior at

x=1

of three functions:

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

,

g(x) = \begin{cases} \frac{x^2-1}{x-1} & x \neq 1 \\ 1 & x = 1 \end{cases}

, and

h(x) = x+1

.

Need a hint?

Simplify

f(x)

and

g(x)

away from

x=1

first, then separately check what happens exactly AT

x=1

for each.

Show solution

1. Step 1: Simplify Away from x=1

For

x \neq 1

,

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

. So all three functions agree everywhere except possibly at

x=1

itself.

2. Step 2: Check the Limit (Same for All Three)

Since all three behave like

x+1

near (but not at)

x=1

:

\lim_{x \to 1} f(x) = \lim_{x \to 1} g(x) = \lim_{x \to 1} h(x) = 2

.

3. Step 3: Check the Actual Value at x=1

f(1)

is undefined.

g(1) = 1

(defined, but doesn't match the limit).

h(1) = 2

(matches the limit).

4. Step 4: Conclude

All three share the exact same limit (2), because a limit only depends on values near

a

, never on the value at

a

. Only

h

happens to also be continuous there.

Practice 02Hard

Determine if

\lim_{x \to 0} \sin\left(\frac{\pi}{x}\right)

exists.

Need a hint?

Think about how many times

\sin(\pi/x)

completes a full oscillation as

x

gets closer and closer to 0.

Show solution

1. Step 1: Examine the Inner Expression

As

x \to 0

,

\frac{\pi}{x} \to \pm\infty

, so the angle inside

\sin(\cdot)

grows without bound.

2. Step 2: Think About the Output

As the angle sweeps through larger and larger values,

\sin\left(\frac{\pi}{x}\right)

keeps cycling through every value in

[-1, 1]

, infinitely many times, no matter how close

x

gets to 0.

3. Step 3: Conclude

Because the function never settles down to a single value,

\lim_{x \to 0} \sin\left(\frac{\pi}{x}\right)

does not exist (DNE) — this is an oscillating limit.

Common Pitfalls

⚠
The 'Value' ConfusionA limit tells you where the function is heading, not where it is. A function can have a limit at a point where it is undefined, or even where it's defined to something completely different (see Examples 1 and 3).
⚠
One Side Isn't EnoughChecking only the left or only the right side is not enough to claim a two-sided limit exists — both must be checked and must agree.
⚠
Oscillation Also Means DNEA limit can fail to exist not just from a left/right mismatch, but also because the function oscillates infinitely and never settles (see Example 4). You'll tame these with the Squeeze Theorem later this week.

Basic Limit Laws (Sum, Difference, Constant Multiple)

Once a limit exists, we rarely estimate it from a table again — instead we break the expression apart using these laws and substitute directly.

Core Theorem

If

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

and

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

, then:

\lim_{x \to a} c = c

\lim_{x \to a} x = a

\lim_{x \to a} [k \cdot f(x)] = k \cdot L

\lim_{x \to a} [f(x) \pm g(x)] = L \pm M

Watch the TikTok ExplanationLimit Laws and Basic Properties→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Given

\lim_{x \to 2} f(x) = 5

, evaluate

\lim_{x \to 2} [3f(x) - 4x + 7]

.

How to think about it

Key idea: Split the expression using the sum/difference law first, then pull constants out with the constant multiple law.

Full solution

1. Step 1: Split the Limit

\lim_{x \to 2} [3f(x)] - \lim_{x \to 2} [4x] + \lim_{x \to 2} [7]

, using the sum and difference rules.

2. Step 2: Pull Out Constants

3 \cdot \lim_{x \to 2} f(x) - 4 \cdot \lim_{x \to 2} x + \lim_{x \to 2} 7

, using the constant multiple law.

3. Step 3: Substitute Known Values

3(5) - 4(2) + 7 = 15 - 8 + 7

.

4. Step 4: Final Answer

15 - 8 + 7 = 14

.

Step-by-Step SOP

1
Split Along Sums and Differences First
Break the expression into individual limits at the outermost $+$ or $-$ before touching constants.
2
Pull Constants Out, Then Substitute
Move constant multipliers outside the limit, then plug in the known limit values last.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Easy

Given

\lim_{x \to 3} f(x) = 4

, evaluate

\lim_{x \to 3} [2f(x) + 5x - 4]

.

Need a hint?

Same pattern as the first example — split, pull out constants, substitute.

Show solution

1. Step 1: Split and Simplify

2 \cdot \lim_{x \to 3} f(x) + 5 \cdot \lim_{x \to 3} x - \lim_{x \to 3} 4

.

2. Step 2: Substitute

2(4) + 5(3) - 4 = 8 + 15 - 4

.

3. Step 3: Final Answer

8 + 15 - 4 = 19

.

Common Pitfalls

⚠
These Laws Require the Limits to ExistYou can only split a limit into pieces if each individual piece's limit actually exists. If one piece is DNE, you can't apply these laws blindly.

Product, Quotient, and Power Laws

The remaining three limit laws let you handle multiplication, division, and exponents — the quotient law has one important condition to watch for.

Core Theorem

If

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

and

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

, then:

\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M

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

, provided

M \neq 0

\lim_{x \to a} [f(x)]^{\alpha} = L^{\alpha}

Watch the TikTok ExplanationAdvanced Limit Laws→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Given

\lim_{x \to 4} f(x) = 9

and

\lim_{x \to 4} g(x) = 2

, evaluate

\lim_{x \to 4} \frac{\sqrt{f(x)}}{x \cdot g(x)}

.

How to think about it

Key idea: Rewrite the square root as a power of $1/2$ so you can apply the power law, then use the product law on the denominator.

Full solution

1. Step 1: Split Into a Quotient of Limits

Since the denominator's limit is non-zero,

\frac{\lim_{x \to 4} \sqrt{f(x)}}{\lim_{x \to 4} [x \cdot g(x)]}

.

2. Step 2: Apply the Power and Product Laws

\frac{[\lim_{x \to 4} f(x)]^{1/2}}{(\lim_{x \to 4} x) \cdot (\lim_{x \to 4} g(x))}

.

3. Step 3: Substitute Known Values

\frac{9^{1/2}}{4 \cdot 2} = \frac{3}{8}

.

Step-by-Step SOP

1
Check the Quotient Condition First
Before dividing limits, verify the denominator's limit is not zero.
2
Rewrite Roots as Powers
Convert $\sqrt{\cdot}$ or $\sqrt[n]{\cdot}$ into fractional exponents so the power law can be applied directly.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Given

\lim_{x \to 1} f(x) = 25

and

\lim_{x \to 1} g(x) = 2

, evaluate

\lim_{x \to 1} \frac{\sqrt{f(x)}}{g(x) + 3}

.

Need a hint?

The denominator is a sum, so apply the sum law inside the limit first, then divide.

Show solution

1. Step 1: Evaluate Numerator and Denominator Separately

\lim_{x \to 1} \sqrt{f(x)} = \sqrt{25} = 5

, and

\lim_{x \to 1} [g(x) + 3] = 2 + 3 = 5

.

2. Step 2: Divide

\frac{5}{5} = 1

.

Practice 02Hard

Why can't you directly apply the quotient law to evaluate

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

?

Need a hint?

Check the condition attached to the quotient law before using it.

Show solution

1. Step 1: Check the Denominator's Limit

\lim_{x \to 2} (x-2) = 0

, so the quotient law's requirement (

M \neq 0

) fails — you cannot divide the limits directly.

2. Step 2: Work Around It

Factor first:

\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2

for

x \neq 2

.

3. Step 3: Final Answer

\lim_{x \to 2} (x+2) = 4

.

Common Pitfalls

⚠
The Quotient Law Needs $M \neq 0$If the denominator's limit is 0, you must simplify algebraically (factor, rationalize) before you can substitute — see Example 3.
⚠
Power Law with Fractional Exponents $\sqrt{f(x)}$ is $[f(x)]^{1/2}$ — treat roots as fractional powers so the power law applies cleanly.

Direct Substitution and Algebraic Manipulation

For polynomials and rational functions, the limit laws collapse into one shortcut: just plug in — as long as you don't land on a $0/0$ that needs algebra first.

Core Theorem

If

p(x)

is a polynomial,

\lim_{x \to a} p(x) = p(a)

(direct substitution).
If

\frac{P(x)}{Q(x)}

is a rational function and

Q(a) \neq 0

, then

\lim_{x \to a} \frac{P(x)}{Q(x)} = \frac{P(a)}{Q(a)}

.
If direct substitution gives

\frac{0}{0}

, simplify first (factor, combine fractions, rationalize) — the limit may still exist.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Evaluate

\lim_{h \to 0} \frac{(3+h)^2 - 9}{h}

.

How to think about it

Key idea: Direct substitution gives $0/0$ — expand the numerator and simplify before substituting.

Full solution

1. Step 1: Expand the Numerator

(3+h)^2 - 9 = 9 + 6h + h^2 - 9 = 6h + h^2

.

2. Step 2: Factor and Cancel

\frac{6h+h^2}{h} = \frac{h(6+h)}{h} = 6+h

, for

h \neq 0

.

3. Step 3: Substitute

\lim_{h \to 0} (6+h) = 6

.

4. Why This Matters

This exact pattern —

\frac{f(a+h)-f(a)}{h}

— is the definition of the derivative you'll meet in Phase 2. You're already doing derivative calculations without realizing it!

Step-by-Step SOP

1
Try Direct Substitution First
It's the fastest path — always attempt it before reaching for algebra.
2
If You Get 0/0, Simplify
Expand, factor, or combine fractions into a single expression, cancel the problematic factor, then substitute again.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

Evaluate

\lim_{x \to 1} \frac{1}{x-1}\left\{\frac{1}{x+3} - \frac{2}{3x+5}\right\}

.

Need a hint?

Combine the inner fraction into a single fraction first — that's usually where a hidden

(x-1)

factor to cancel is hiding.

Show solution

1. Step 1: Combine the Inner Fractions

\frac{1}{x+3} - \frac{2}{3x+5} = \frac{(3x+5) - 2(x+3)}{(x+3)(3x+5)} = \frac{x-1}{(x+3)(3x+5)}

.

2. Step 2: Multiply by the Outer Factor

\frac{1}{x-1} \cdot \frac{x-1}{(x+3)(3x+5)} = \frac{1}{(x+3)(3x+5)}

, for

x \neq 1

.

3. Step 3: Substitute

\frac{1}{(1+3)(3(1)+5)} = \frac{1}{4 \cdot 8} = \frac{1}{32}

.

Common Pitfalls

⚠
Don't Panic at 0/0Getting $0/0$ from direct substitution doesn't mean the limit is 0 or DNE — it means you have more algebra to do (factor, combine fractions) before you can substitute.
⚠
Combine Before You CancelWhen a complex fraction is involved, combine the inner terms into a single fraction first — the cancelling factor is often hidden inside that combination (see Example 2).

The Formal ε-δ Definition of a Limit

The intuitive 'gets closer and closer' picture from earlier is great for building a feel for limits, but it's not a rigorous mathematical statement. The ε-δ definition makes 'arbitrarily close' precise using two competing challenges — and lets you actually prove a limit is what you claim it is.

Core Theorem

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

means: for every

\varepsilon > 0

, there exists a

\delta > 0

such that whenever

0 < |x-c| < \delta

, it follows that

|f(x)-L| < \varepsilon

.
In words: no matter how small a tolerance

\varepsilon

someone challenges you with around

L

, you can always find a window

\delta

around

c

(excluding

c

itself) that keeps

f(x)

trapped inside that tolerance.
The same idea extends to one-sided limits (restrict to

c < x < c+\delta

or

c-\delta < x < c

), limits at infinity (replace

\delta

with 'for all

x > M

'), and infinite limits (replace

\varepsilon

with 'for all

B > 0

,

f(x) > B

').

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Hard

Use the ε-δ definition to prove

\lim_{x \to 3} (4x-5) = 7

.

How to think about it

Key idea: Start from the target inequality $|(4x-5)-7| < \varepsilon$ and simplify until you can see exactly how $\delta$ should depend on $\varepsilon$ .

Full solution

1. Step 1: Simplify the Target Inequality

We want

|(4x-5)-7| < \varepsilon

. This simplifies to

|4x-12| < \varepsilon

, i.e.

4|x-3| < \varepsilon

.

2. Step 2: Solve for |x-3|

Dividing by 4:

|x-3| < \frac{\varepsilon}{4}

. This tells us exactly how close

x

needs to be to 3.

3. Step 3: Choose δ

Let

\delta = \frac{\varepsilon}{4}

. Then whenever

0 < |x-3| < \delta

, we get

|(4x-5)-7| = 4|x-3| < 4\delta = \varepsilon

.

4. Step 4: Conclude

Since we found a valid

\delta

for every

\varepsilon > 0

, this proves

\lim_{x \to 3} (4x-5) = 7

.

\blacksquare

Step-by-Step SOP

1
Start From |f(x) - L| < ε
Write out the target inequality and simplify/factor it in terms of $|x-c|$ .
2
Bound Any Non-Constant Factor (If Needed)
If a variable factor remains besides $|x-c|$ , temporarily restrict $\delta \le 1$ to pin down a numeric bound for it.
3
Solve for δ and Take the Minimum
Isolate $|x-c|$ to see what $\delta$ needs to be, taking the minimum of all constraints you generated.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

Use the ε-δ definition to prove

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

.

Need a hint?

Unlike the linear case,

|x^2-9| = |x-3||x+3|

has a second factor that isn't constant — you'll need to first restrict

\delta \le 1

to bound it.

Show solution

1. Step 1: Factor the Target Inequality

We want

|x^2-9| < \varepsilon

. Factor:

|x^2-9| = |x-3||x+3|

.

2. Step 2: Bound the Extra Factor

Temporarily assume

\delta \le 1

, so

|x-3| < 1 \implies 2 < x < 4 \implies 5 < x+3 < 7

. This means

|x+3| < 7

.

3. Step 3: Combine the Bounds

Now

|x^2-9| = |x-3||x+3| < 7|x-3|

. We want

7|x-3| < \varepsilon

, i.e.

|x-3| < \frac{\varepsilon}{7}

.

4. Step 4: Choose δ

Let

\delta = \min\left(1, \frac{\varepsilon}{7}\right)

— the smaller of the two requirements. Then

0 < |x-3| < \delta \implies |x^2-9| < 7 \cdot \frac{\varepsilon}{7} = \varepsilon

.

5. Step 5: Conclude

Since a valid

\delta

exists for every

\varepsilon > 0

, this proves

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

.

\blacksquare

Common Pitfalls

⚠
δ Depends on ε, Never the ReverseYou always start by assuming $\varepsilon$ is given (fixed but arbitrary), then hunt for a $\delta$ in terms of it — never the other way around.
⚠
Non-Linear Functions Need a Bounding StepWhen $f(x)-L$ factors into $(x-c)$ times something non-constant (like Example 2's $|x+3|$ ), you must first cap $\delta$ (commonly $\delta \le 1$ ) to bound that extra factor before solving for the final $\delta$ .
⚠
Take the MinimumIf you generated two separate requirements on $\delta$ (one from the bounding step, one from the main inequality), your final answer is $\delta = \min(\cdot, \cdot)$ — it must satisfy both simultaneously.

Worked Examples

1. Step 1: Read the Left Side

2. Step 2: Read the Right Side

3. Step 3: Conclude

1. Step 1: Test the Right Side

2. Step 2: Test the Left Side

3. Step 3: Compare and Conclude

Check Left and Right Separately

Compare, Then Conclude

Practice — Your Turn

1. Step 1: Simplify Away from x=1

2. Step 2: Check the Limit (Same for All Three)

3. Step 3: Check the Actual Value at x=1

4. Step 4: Conclude

1. Step 1: Examine the Inner Expression

2. Step 2: Think About the Output

3. Step 3: Conclude

Worked Examples

1. Step 1: Split the Limit

2. Step 2: Pull Out Constants

3. Step 3: Substitute Known Values

4. Step 4: Final Answer

Split Along Sums and Differences First

Pull Constants Out, Then Substitute

Practice — Your Turn

1. Step 1: Split and Simplify

2. Step 2: Substitute

3. Step 3: Final Answer

Worked Examples

1. Step 1: Split Into a Quotient of Limits

2. Step 2: Apply the Power and Product Laws

3. Step 3: Substitute Known Values

Check the Quotient Condition First

Rewrite Roots as Powers

Practice — Your Turn

1. Step 1: Evaluate Numerator and Denominator Separately

2. Step 2: Divide

1. Step 1: Check the Denominator's Limit

2. Step 2: Work Around It

3. Step 3: Final Answer

Worked Examples

1. Step 1: Expand the Numerator

2. Step 2: Factor and Cancel

3. Step 3: Substitute

4. Why This Matters

Try Direct Substitution First

If You Get 0/0, Simplify

Practice — Your Turn

1. Step 1: Combine the Inner Fractions

2. Step 2: Multiply by the Outer Factor

3. Step 3: Substitute

Worked Examples

1. Step 1: Simplify the Target Inequality

2. Step 2: Solve for |x-3|

3. Step 3: Choose δ

4. Step 4: Conclude

Start From |f(x) - L| < ε

Bound Any Non-Constant Factor (If Needed)

Solve for δ and Take the Minimum

Practice — Your Turn

1. Step 1: Factor the Target Inequality

2. Step 2: Bound the Extra Factor

3. Step 3: Combine the Bounds

4. Step 4: Choose δ

5. Step 5: Conclude

Gary Chang