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Direct Substitution Method

The simplest limit technique: when you can just plug in the value

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The Direct Substitution Method

The easiest way to find a limit? Just plug in the number!

When It Works

Direct substitution works when the function is continuous at the point. This includes:

  • Polynomial functions: x2x^2, 3x3−2x+13x^3 - 2x + 1, etc.
  • Rational functions (when denominator ≠ 0)
  • Radical functions (when defined)
  • Trigonometric functions (at most points)
  • Exponential and logarithmic functions

If you can evaluate f(a)f(a) without any problems, then lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a)

The Process

To find lim⁡x→af(x)\lim_{x \to a} f(x):

  1. Try substituting x = a directly into f(x)
  2. If you get a number → That's your answer! ✓
  3. If you get 00\frac{0}{0}, ∞∞\frac{\infty}{\infty}, etc. → Need a different technique

Example 1: Polynomial

Find lim⁡x→2(x2+3x−1)\lim_{x \to 2} (x^2 + 3x - 1)

Solution: Just substitute x = 2:

lim⁡x→2(x2+3x−1)=(2)2+3(2)−1=4+6−1=9\lim_{x \to 2} (x^2 + 3x - 1) = (2)^2 + 3(2) - 1 = 4 + 6 - 1 = 9

✓ Answer: 9

Example 2: Rational Function (Works)

Find lim⁡x→3x+1x−1\lim_{x \to 3} \frac{x + 1}{x - 1}

Solution: Substitute x = 3:

lim⁡x→3x+1x−1=3+13−1=42=2\lim_{x \to 3} \frac{x + 1}{x - 1} = \frac{3 + 1}{3 - 1} = \frac{4}{2} = 2

✓ Answer: 2

Example 3: When It Fails

Find lim⁡x→1x−1x2−1\lim_{x \to 1} \frac{x - 1}{x^2 - 1}

Attempt: 1−112−1=00\frac{1 - 1}{1^2 - 1} = \frac{0}{0}

✗ This is indeterminate! Direct substitution doesn't work here.

When you get 00\frac{0}{0}, you need algebraic manipulation (factoring, rationalizing, etc.)

Indeterminate Forms

These special cases mean direct substitution has failed:

FormWhat to Do
00\frac{0}{0}Factor, simplify, rationalize
∞∞\frac{\infty}{\infty}Divide by highest power
0⋅∞0 \cdot \inftyRewrite as a fraction
∞−∞\infty - \inftyCombine terms differently

We'll cover techniques for these in upcoming lessons!

Quick Check Method

Ask yourself: "Can I safely plug in this number?"

  • Is the function defined there? → Try it!
  • Does the denominator become zero? → Can't use direct substitution
  • Is there a square root of a negative? → Can't use direct substitution
  • Otherwise → Go for it!

Practice Tip

Always try direct substitution first. It's the fastest method when it works!

If you get a real number, you're done. If you get an indeterminate form, move on to other techniques.

📚 Practice Problems

1Problem 1easy

❓ Question:

Evaluate lim⁡x→4(2x2−5x+3)\lim_{x \to 4} (2x^2 - 5x + 3) using direct substitution.

💡 Show Solution

Since this is a polynomial, we can substitute directly:

lim⁡x→4(2x2−5x+3)\lim_{x \to 4} (2x^2 - 5x + 3)

Substitute x = 4:

=2(4)2−5(4)+3= 2(4)^2 - 5(4) + 3 =2(16)−20+3= 2(16) - 20 + 3 =32−20+3= 32 - 20 + 3 =15= 15

Answer: 15

2Problem 2easy

❓ Question:

Evaluate the following limits using direct substitution:

a) lim⁡x→3(2x2−5x+1)\lim_{x \to 3} (2x^2 - 5x + 1) b) lim⁡x→−2x3+8x+2\lim_{x \to -2} \frac{x^3 + 8}{x + 2} c) lim⁡x→0sin⁡xx\lim_{x \to 0} \frac{\sin x}{x}

💡 Show Solution

Solution:

Part (a): The function is a polynomial, which is continuous everywhere.

Direct substitution:

lim⁡x→3(2x2−5x+1)=2(3)2−5(3)+1=18−15+1=4\lim_{x \to 3} (2x^2 - 5x + 1) = 2(3)^2 - 5(3) + 1 = 18 - 15 + 1 = 4

Part (b): Direct substitution gives 00\frac{0}{0} (indeterminate).

Factor the numerator (sum of cubes): x3+8=(x+2)(x2−2x+4)x^3 + 8 = (x+2)(x^2-2x+4)

lim⁡x→−2(x+2)(x2−2x+4)x+2=lim⁡x→−2(x2−2x+4)\lim_{x \to -2} \frac{(x+2)(x^2-2x+4)}{x+2} = \lim_{x \to -2} (x^2-2x+4)

Now use direct substitution:

=(−2)2−2(−2)+4=4+4+4=12= (-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12

Part (c): This is a special limit that cannot be evaluated by direct substitution (gives 00\frac{0}{0}).

This is a fundamental limit in calculus:

lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1

This must be memorized or proven using the squeeze theorem.

3Problem 3medium

❓ Question:

Try to evaluate lim⁡x→5x2−25x−5\lim_{x \to 5} \frac{x^2 - 25}{x - 5} using direct substitution. What happens?

💡 Show Solution

Let's try substituting x = 5:

lim⁡x→5x2−25x−5=52−255−5=25−250=00\lim_{x \to 5} \frac{x^2 - 25}{x - 5} = \frac{5^2 - 25}{5 - 5} = \frac{25 - 25}{0} = \frac{0}{0}

Result: Indeterminate form 00\frac{0}{0}

Direct substitution fails because we get an indeterminate form.

To actually solve this, we need to factor the numerator:

lim⁡x→5x2−25x−5=lim⁡x→5(x−5)(x+5)x−5=lim⁡x→5(x+5)=10\lim_{x \to 5} \frac{x^2 - 25}{x - 5} = \lim_{x \to 5} \frac{(x-5)(x+5)}{x - 5} = \lim_{x \to 5} (x + 5) = 10

The limit exists and equals 10, but we couldn't find it with direct substitution alone.

4Problem 4hard

❓ Question:

Evaluate lim(x→2) (x³ - 2x² + 5x - 1)

💡 Show Solution

Step 1: Check if function is continuous at x = 2: This is a polynomial, which is continuous everywhere

Step 2: Apply direct substitution: Simply substitute x = 2 into the function

Step 3: Calculate: (2)³ - 2(2)² + 5(2) - 1 = 8 - 8 + 10 - 1 = 9

Step 4: Conclusion: lim(x→2) (x³ - 2x² + 5x - 1) = 9

Answer: 9

Explain using:

📋 AP Calculus AB — Exam Format Guide

⏱ 3 hours 15 minutes📝 51 questions📊 4 sections
SectionFormatQuestionsTimeWeightCalculator
Multiple Choice (No Calculator)MCQ3060 min33.3%🚫
Multiple Choice (Calculator)MCQ1545 min16.7%✅
Free Response (Calculator)FRQ230 min16.7%✅
Free Response (No Calculator)FRQ460 min33.3%🚫

📊 Scoring: 1-5

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💡 Key Test-Day Tips

  • ✓Show all work on FRQs
  • ✓Use proper notation
  • ✓Check units
  • ✓Manage your time

⚠️ Common Mistakes: Direct Substitution Method

Avoid these 4 frequent errors

🌍 Real-World Applications: Direct Substitution Method

See how this math is used in the real world

📝 Worked Example: Related Rates — Expanding Circle

Problem:

A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 22 cm/s. How fast is the area of the circle increasing when the radius is 1010 cm?

2Write the relationship between variables
3Differentiate both sides with respect to time
4Substitute known values

📌 Related Topics in Limits & Continuity

❓ Frequently Asked Questions

What is Direct Substitution Method?▾
The simplest limit technique: when you can just plug in the value
How can I study Direct Substitution Method effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 4 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Direct Substitution Method study guide free?▾
Yes — all study notes, flashcards, and practice problems for Direct Substitution Method on Study Mondo are free to access. No account is needed.
What course covers Direct Substitution Method?▾
Direct Substitution Method is part of the AP Calculus AB course on Study Mondo, specifically in the Limits & Continuity section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Direct Substitution Method?▾
Yes, this page includes 4 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.