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.
Direct Substitution
lim[x->a] f(x) = f(a)
To find the limit, just plug in x = a and calculate.
Find lim[x->4] (x^2 - 3x + 2)
1
Try direct substitution
Plug in x = 4:
(4)^2 - 3(4) + 2 = 16 - 12 + 2 = 6
2
Check: Did we get a real number?
Yes! We got 6, which is a normal number. No division by zero, no weirdness.
3
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."
Find lim[x->2] (x^2 - 4)/(x - 2)
1
Try direct substitution
Plug in x = 2:
(2^2 - 4)/(2 - 2) = (4 - 4)/(0) = **0/0** <- INDETERMINATE!
2
Don't panic! Factor the numerator
x^2 - 4 is a difference of squares:
x^2 - 4 = (x + 2)(x - 2)
3
Rewrite the expression
(x^2 - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2)
4
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.*
5
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)
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
Always try direct substitution first - it's quick and often works
If you get a number, you're done
If you get 0/0, you need to simplify and try again
Common fix: factor and cancel common terms