Direct Substitution Method

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

The Process

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

1. Try substituting x = a directly into f(x)
2. If you get a number → That's your answer! ✓
3. 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:

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.

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.

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.