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Direct Substitution & When It Fails

🔉 80%

In the last lesson, we found limits by simply plugging in the value. That's called direct substitution, and it's always your first move.

But here's the thing: it doesn't always work. Sometimes plugging in gives you something weird like 0/0. That's when the real fun begins.

Function with a Hole

When direct substitution fails - the function has a hole but still approaches a limit

Illustration: function-with-hole

Direct Substitution

lim[x->a] f(x) = f(a)

To find the limit, just plug in x = a and calculate.

When Direct Substitution Works

Easy

Find lim[x->4] (x^2 - 3x + 2)

Try direct substitution

Plug in x = 4: (4)^2 - 3(4) + 2 = 16 - 12 + 2 = 6

Check: Did we get a real number?

Yes! We got 6, which is a normal number. No division by zero, no weirdness.

Write the answer

lim[x->4] (x^2 - 3x + 2) = 6

Answer: 6

⚠️

The 0/0 Problem

What happens when direct substitution gives you 0/0?

This is called an indeterminate form. It doesn't mean the limit doesn't exist - it means we need to work harder to find it.

Think of 0/0 as calculus saying: "I can't tell yet. Give me more information."

When Direct Substitution FAILS

Medium

Find lim[x->2] (x^2 - 4)/(x - 2)

Try direct substitution

Plug in x = 2: (2^2 - 4)/(2 - 2) = (4 - 4)/(0) = **0/0** <- INDETERMINATE!

Don't panic! Factor the numerator

x^2 - 4 is a difference of squares: x^2 - 4 = (x + 2)(x - 2)

Rewrite the expression

(x^2 - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2)

Cancel the common factor

(x + 2)(x - 2)/(x - 2) = (x + 2) *Note: We can cancel because we're looking at the limit as x APPROACHES 2, not AT x = 2.*

Now use direct substitution

lim[x->2] (x + 2) = 2 + 2 = 4

Answer: 4

💡 The 0/0 was hiding a common factor! Once we factored and canceled, the limit became easy.

1. Find lim[x->3] (x^2 - 9)/(x - 3)

Does not exist

x^2 - 9 = (x+3)(x-3), so (x^2-9)/(x-3) = (x+3)(x-3)/(x-3) = x+3. As x->3, x+3->6.

Key Takeaways