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Introduction
1. Limits
1.1 Tables and graphs
1.2 Analytical limits
1.2.1 Direct substitution
1.2.2 Composite functions
1.3 Algebraic limits
1.4 Limits and infinity
1.5 Special limits
1.6 Continuity
2. Derivative basics
3. Advanced differentiation
4. Contextual uses
5. Analytical uses
6. Integration
7. Differential equations
8. Applications of integrals
Wrapping up
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1.2.1 Direct substitution
Achievable AP Calculus AB
1. Limits
1.2. Analytical limits
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Direct substitution

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What you’ll learn

  • Direct substitution: Evaluate limit values by direct algebraic substitution.
  • Limit laws: Apply fundamental limit properties to simplify and evaluate expressions.
  • Graphical interpretation: Combine algebraic limit laws with visual data from graphs.

Evaluating limits analytically

While creating graphs and tables are excellent tools for visualizing behaviors of functions or double-checking your work, most limits can be evaluated analytically using algebra and calculus. For many functions, straightforward techniques such as direct substitution and the basic limit laws can be used to determine a limit.

Direct substitution

The simplest way to evaluate a limit is direct substitution. If plugging in x=a into f(x) gives a finite number, then that number is the limit.

AP tip:

Always try direct substitution first.

Example 1. Evaluate

x→−3lim​(x2+x)

(spoiler)

Substituting x=−3 into the expression directly,

x→−3lim​(x2+x)​=(−3)2+(−3)=9−3=6​

Example 2. Evaluate

x→5lim​x−1​x+1​

(spoiler)

Substituting x=5 directly,

x→5lim​x−1​x+1​​=5−1​5+1​=26​=3​

Limit properties

A complicated limit can be broken down into smaller pieces by calculating the limit of each part individually. Limit laws follow standard mathematical rules.

Use these when a specific equation is not given and you must work with graphs or limit values instead.

If we know that

x→alim​f(x)=Landx→alim​g(x)=M

then

Property Limit law
Constant (c is a constant) x→alim​c=c
Identity x→alim​x=a
Constant multiple x→alim​[c×f(x)]=c×L
Sum/difference x→alim​[f(x)±g(x)]=L±M
Product x→alim​[f(x)×g(x)]=L×M
Quotient x→alim​g(x)f(x)​=ML​ (where M=0)
Power (constant n) x→alim​[f(x)]n=Ln

Warning: Inside vs. outside constants

The constant multiple rule only applies when a constant is multiplying the entire function on the outside. You cannot pull a constant out of a function’s input variable.

x→2lim​f(3x)=3⋅x→2lim​f(x)

Instead, plug the x-value into the inside expression first to find your actual target on the graph:

x→2lim​f(3x)⇒Find the limit of f(x) as x approaches 6.

Examples

Example 1.

Given x→5lim​f(x)=2 and x→5lim​g(x)=4, evaluate:

x→5lim​ [f(x)−3g(x)]2

(spoiler)

Applying limit laws,

x→5lim​ [f(x)−3g(x)]2=(x→5lim​f(x)−3⋅x→5lim​g(x))2=(2−3⋅4)2=100

Example 2.

If x→−2lim​f(x)=−3 and x→−2lim​g(x)=6, evaluate:

x→−2lim​f(x)2g(x)−f(x)+7​​

(spoiler)

The power law also handles roots since x​=x1/2.

Applying limit laws,

x→−2lim​f(x)2g(x)−f(x)+7​​=(x→−2lim​f(x))2x→−2lim​g(x)−x→−2lim​f(x)+x→−2lim​7​​=(−3)26−(−3)+7​​=916​​=34​

Using graphs

Limit laws can also be used after determining limit values from a graph when functions are not explicitly given.

Use the following graphs of f(x) and g(x) to determine the limits below.

Applying limit laws
Applying limit laws

a) x→2lim​[x2⋅f(x)]

b) x→−1lim​f(x−1)

c) x→1lim​[g(2x)+2g(x)]−g(2)

Answers

(spoiler)

a) 12
b) 3
c) 2

Solutions

a) x→2lim​[x2⋅f(x)]

(spoiler)

Applying limit laws:

x→2lim​[x2⋅f(x)]​=(x→2lim​x2)⋅(x→2lim​f(x))=(22)⋅3=12​

b) x→−1lim​f(x−1)

(spoiler)

As x approaches −1, the input x−1 approaches

−1−1=−2

So we find the limit of f(x) as x→−2. From the graph, this limit is 3.

c) x→1lim​[g(2x)+2g(x)]−g(2)

(spoiler)

Apply limit laws to distribute the limit across the terms:

[x→1lim​g(2x)+2⋅x→1lim​g(x)]−g(2)

Note: The limit operator only applies to the terms inside the brackets. Because g(2) is outside the brackets, it is evaluated separately.

Now evaluate each piece using the graph:

  • 1st term: x→1lim​g(2x)=1

As x approaches 1, the input 2x approaches 2. So we find the limit of g(x) as x→2. From the graph, this limit is 1.

  • 2nd term: 2⋅x→1lim​g(x)=0

Since x→1lim​g(x)=0, then 2⋅x→1lim​g(x)=0.

  • 3rd term: g(2)=−1

This is a function value, not a limit. Look strictly at the solid dot at x=2, which is at y=−1.

Substitute these values back into the expression:

[x→1lim​g(2x)+2⋅x→1lim​g(x)]−g(2)=[1+0]−(−1)=2

Direct substitution

  • Simplest method: plug x=a directly into f(x)
  • If result is a finite number, that is the limit
  • Always try this first

Limit laws

  • Break complex limits into smaller parts evaluated individually
  • Key properties (given limx→a​f(x)=L, limx→a​g(x)=M):
    • Constant multiple: lim[c⋅f(x)]=c⋅L
    • Sum/difference: lim[f(x)±g(x)]=L±M
    • Product: lim[f(x)⋅g(x)]=L⋅M
    • Quotient: limg(x)f(x)​=ML​, where M=0
    • Power: lim[f(x)]n=Ln (also applies to roots, e.g., n=21​)

Graphical interpretation

  • Read limit values of f(x) and g(x) directly from graphs
  • Apply limit laws using those values to evaluate combined expressions

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Direct substitution

What you’ll learn

  • Direct substitution: Evaluate limit values by direct algebraic substitution.
  • Limit laws: Apply fundamental limit properties to simplify and evaluate expressions.
  • Graphical interpretation: Combine algebraic limit laws with visual data from graphs.

Evaluating limits analytically

While creating graphs and tables are excellent tools for visualizing behaviors of functions or double-checking your work, most limits can be evaluated analytically using algebra and calculus. For many functions, straightforward techniques such as direct substitution and the basic limit laws can be used to determine a limit.

Direct substitution

The simplest way to evaluate a limit is direct substitution. If plugging in x=a into f(x) gives a finite number, then that number is the limit.

AP tip:

Always try direct substitution first.

Example 1. Evaluate

x→−3lim​(x2+x)

(spoiler)

Substituting x=−3 into the expression directly,

x→−3lim​(x2+x)​=(−3)2+(−3)=9−3=6​

Example 2. Evaluate

x→5lim​x−1​x+1​

(spoiler)

Substituting x=5 directly,

x→5lim​x−1​x+1​​=5−1​5+1​=26​=3​

Limit properties

A complicated limit can be broken down into smaller pieces by calculating the limit of each part individually. Limit laws follow standard mathematical rules.

Use these when a specific equation is not given and you must work with graphs or limit values instead.

If we know that

x→alim​f(x)=Landx→alim​g(x)=M

then

Property Limit law
Constant (c is a constant) x→alim​c=c
Identity x→alim​x=a
Constant multiple x→alim​[c×f(x)]=c×L
Sum/difference x→alim​[f(x)±g(x)]=L±M
Product x→alim​[f(x)×g(x)]=L×M
Quotient x→alim​g(x)f(x)​=ML​ (where M=0)
Power (constant n) x→alim​[f(x)]n=Ln

Warning: Inside vs. outside constants

The constant multiple rule only applies when a constant is multiplying the entire function on the outside. You cannot pull a constant out of a function’s input variable.

x→2lim​f(3x)=3⋅x→2lim​f(x)

Instead, plug the x-value into the inside expression first to find your actual target on the graph:

x→2lim​f(3x)⇒Find the limit of f(x) as x approaches 6.

Examples

Example 1.

Given x→5lim​f(x)=2 and x→5lim​g(x)=4, evaluate:

x→5lim​ [f(x)−3g(x)]2

(spoiler)

Applying limit laws,

x→5lim​ [f(x)−3g(x)]2=(x→5lim​f(x)−3⋅x→5lim​g(x))2=(2−3⋅4)2=100

Example 2.

If x→−2lim​f(x)=−3 and x→−2lim​g(x)=6, evaluate:

x→−2lim​f(x)2g(x)−f(x)+7​​

(spoiler)

The power law also handles roots since x​=x1/2.

Applying limit laws,

x→−2lim​f(x)2g(x)−f(x)+7​​=(x→−2lim​f(x))2x→−2lim​g(x)−x→−2lim​f(x)+x→−2lim​7​​=(−3)26−(−3)+7​​=916​​=34​

Using graphs

Limit laws can also be used after determining limit values from a graph when functions are not explicitly given.

Use the following graphs of f(x) and g(x) to determine the limits below.

a) x→2lim​[x2⋅f(x)]

b) x→−1lim​f(x−1)

c) x→1lim​[g(2x)+2g(x)]−g(2)

Answers

(spoiler)

a) 12
b) 3
c) 2

Solutions

a) x→2lim​[x2⋅f(x)]

(spoiler)

Applying limit laws:

x→2lim​[x2⋅f(x)]​=(x→2lim​x2)⋅(x→2lim​f(x))=(22)⋅3=12​

b) x→−1lim​f(x−1)

(spoiler)

As x approaches −1, the input x−1 approaches

−1−1=−2

So we find the limit of f(x) as x→−2. From the graph, this limit is 3.

c) x→1lim​[g(2x)+2g(x)]−g(2)

(spoiler)

Apply limit laws to distribute the limit across the terms:

[x→1lim​g(2x)+2⋅x→1lim​g(x)]−g(2)

Note: The limit operator only applies to the terms inside the brackets. Because g(2) is outside the brackets, it is evaluated separately.

Now evaluate each piece using the graph:

  • 1st term: x→1lim​g(2x)=1

As x approaches 1, the input 2x approaches 2. So we find the limit of g(x) as x→2. From the graph, this limit is 1.

  • 2nd term: 2⋅x→1lim​g(x)=0

Since x→1lim​g(x)=0, then 2⋅x→1lim​g(x)=0.

  • 3rd term: g(2)=−1

This is a function value, not a limit. Look strictly at the solid dot at x=2, which is at y=−1.

Substitute these values back into the expression:

[x→1lim​g(2x)+2⋅x→1lim​g(x)]−g(2)=[1+0]−(−1)=2

Key points

Direct substitution

  • Simplest method: plug x=a directly into f(x)
  • If result is a finite number, that is the limit
  • Always try this first

Limit laws

  • Break complex limits into smaller parts evaluated individually
  • Key properties (given limx→a​f(x)=L, limx→a​g(x)=M):
    • Constant multiple: lim[c⋅f(x)]=c⋅L
    • Sum/difference: lim[f(x)±g(x)]=L±M
    • Product: lim[f(x)⋅g(x)]=L⋅M
    • Quotient: limg(x)f(x)​=ML​, where M=0
    • Power: lim[f(x)]n=Ln (also applies to roots, e.g., n=21​)

Graphical interpretation

  • Read limit values of f(x) and g(x) directly from graphs
  • Apply limit laws using those values to evaluate combined expressions

More from Analytical limits

  • Composite functions