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)

.

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 Exercises

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

NEED A HINT?

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.

SHOW DETAILED EXPLANATION

Step 1: Read the Left Side

As

x \to 1^-

(0.9, 0.99, 0.999),

f(x)

is approaching 2.

Step 2: Read the Right Side

As

x \to 1^+

(1.1, 1.01, 1.001),

f(x)

is also approaching 2.

Step 3: Conclude

Since both sides approach the same value,

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

, even though

f(1)

itself is undefined.

Example 02Medium

Determine if

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

exists.

NEED A HINT?

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

SHOW DETAILED EXPLANATION

Step 1: Test the Right Side

For

x > 3

,

|x-3| = x-3

. Therefore,

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

.

Step 2: Test the Left Side

For

x < 3

,

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

. Therefore,

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

.

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)**.

Example 03Medium

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 DETAILED EXPLANATION

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.

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

.

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).

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.

Example 04Hard

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 DETAILED EXPLANATION

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.

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.

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→

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 Exercises

Example 01Easy

Given

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

, evaluate

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

.

Watch the TikTok ExplanationLimit Laws and Basic Properties→

NEED A HINT?

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

SHOW DETAILED EXPLANATION

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.

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.

Step 3: Substitute Known Values

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

.

Step 4: Final Answer

15 - 8 + 7 = 14

.

Example 02Easy

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 DETAILED EXPLANATION

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

.

Step 2: Substitute

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

.

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→

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 Exercises

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)}

.

Watch the TikTok ExplanationAdvanced Limit Laws→

NEED A HINT?

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.

SHOW DETAILED EXPLANATION

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)]}

.

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))}

.

Step 3: Substitute Known Values

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

.

Example 02Medium

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 DETAILED EXPLANATION

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

.

Step 2: Divide

\frac{5}{5} = 1

.

Example 03Hard

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 DETAILED EXPLANATION

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.

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

.

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.

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 Exercises

Example 01Medium

Evaluate

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

.

NEED A HINT?

Direct substitution gives

0/0

— expand the numerator and simplify before substituting.

SHOW DETAILED EXPLANATION

Step 1: Expand the Numerator

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

.

Step 2: Factor and Cancel

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

, for

h \neq 0

.

Step 3: Substitute

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

.

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!

Example 02Hard

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 DETAILED EXPLANATION

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)}

.

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

.

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).

Check Left and Right Separately

Compare, Then Conclude

Practice Exercises

Step 1: Read the Left Side

Step 2: Read the Right Side

Step 3: Conclude

Step 1: Test the Right Side

Step 2: Test the Left Side

Step 3: Compare and Conclude

Step 1: Simplify Away from x=1

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

Step 3: Check the Actual Value at x=1

Step 4: Conclude

Step 1: Examine the Inner Expression

Step 2: Think About the Output

Step 3: Conclude

Split Along Sums and Differences First

Pull Constants Out, Then Substitute

Practice Exercises

Step 1: Split the Limit

Step 2: Pull Out Constants

Step 3: Substitute Known Values

Step 4: Final Answer

Step 1: Split and Simplify

Step 2: Substitute

Step 3: Final Answer

Check the Quotient Condition First

Rewrite Roots as Powers

Practice Exercises

Step 1: Split Into a Quotient of Limits

Step 2: Apply the Power and Product Laws

Step 3: Substitute Known Values

Step 1: Evaluate Numerator and Denominator Separately

Step 2: Divide

Step 1: Check the Denominator's Limit

Step 2: Work Around It

Step 3: Final Answer

Try Direct Substitution First

If You Get 0/0, Simplify

Practice Exercises

Step 1: Expand the Numerator

Step 2: Factor and Cancel

Step 3: Substitute

Why This Matters

Step 1: Combine the Inner Fractions

Step 2: Multiply by the Outer Factor

Step 3: Substitute

Gary Chang