Direct Substitution Method

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

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

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

The Process

To find $\lim_{x \to a} f(x)$ :

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

Example 1: Polynomial

Find $\lim_{x \to 2} (x^2 + 3x - 1)$

Solution: Just substitute x = 2:

$\lim_{x \to 2} (x^2 + 3x - 1) = (2)^2 + 3(2) - 1 = 4 + 6 - 1 = 9$

✓ Answer: 9

Example 2: Rational Function (Works)

Find $\lim_{x \to 3} \frac{x + 1}{x - 1}$

Solution: Substitute x = 3:

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

✓ Answer: 2

Example 3: When It Fails

Find $\lim_{x \to 1} \frac{x - 1}{x^2 - 1}$

Attempt: $\frac{1 - 1}{1^2 - 1} = \frac{0}{0}$

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

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

Indeterminate Forms

These special cases mean direct substitution has failed:

| Form | What to Do | |------|------------| | $\frac{0}{0}$ | Factor, simplify, rationalize | | $\frac{\infty}{\infty}$ | Divide by highest power | | $0 \cdot \infty$ | Rewrite as a fraction | | $\infty - \infty$ | Combine 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 $\lim_{x \to 4} (2x^2 - 5x + 3)$ using direct substitution.

💡 Show Solution

Since this is a polynomial, we can substitute directly:

$\lim_{x \to 4} (2x^2 - 5x + 3)$

Substitute x = 4:

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

Answer: 15

2Problem 2easy

❓ Question:

Evaluate the following limits using direct substitution:

a) $\lim_{x \to 3} (2x^2 - 5x + 1)$ b) $\lim_{x \to -2} \frac{x^3 + 8}{x + 2}$ c) $\lim_{x \to 0} \frac{\sin x}{x}$

💡 Show Solution

Solution:

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

Direct substitution:

$\lim_{x \to 3} (2x^2 - 5x + 1) = 2(3)^2 - 5(3) + 1 = 18 - 15 + 1 = 4$

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

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

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

Now use direct substitution:

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

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

This is a fundamental limit in calculus:

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

This must be memorized or proven using the squeeze theorem.

3Problem 3easy

❓ Question:

Evaluate the following limits using direct substitution:

a) $\lim_{x \to 3} (2x^2 - 5x + 1)$ b) $\lim_{x \to -2} \frac{x^3 + 8}{x + 2}$ c) $\lim_{x \to 0} \frac{\sin x}{x}$

💡 Show Solution

Solution:

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

Direct substitution:

$\lim_{x \to 3} (2x^2 - 5x + 1) = 2(3)^2 - 5(3) + 1 = 18 - 15 + 1 = 4$

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

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

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

Now use direct substitution:

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

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

This is a fundamental limit in calculus:

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

This must be memorized or proven using the squeeze theorem.

4Problem 4medium

❓ Question:

Try to evaluate $\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$ using direct substitution. What happens?

💡 Show Solution

Let's try substituting x = 5:

$\lim_{x \to 5} \frac{x^2 - 25}{x - 5} = \frac{5^2 - 25}{5 - 5} = \frac{25 - 25}{0} = \frac{0}{0}$

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

Direct substitution fails because we get an indeterminate form.

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

$\lim_{x \to 5} \frac{x^2 - 25}{x - 5} = \lim_{x \to 5} \frac{(x-5)(x+5)}{x - 5} = \lim_{x \to 5} (x + 5) = 10$

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

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