A limit asks where the function is heading, not where it is. Plug in first — and when 00 appears, that is not an answer, it is an instruction.
The problem
limx→2 x² − 4x − 2
Plugging in 2 gives 00 — a signal to factor: (x − 2)x + 2x − 2 = x + 2 away from 2, and x + 2 → 4. "Undefined" answers a different question: f(2) does not exist, but the limit — where the heights are heading — does.
Most limits on exams are one substitution. Only special cases need more.
2² − 42 − 2 = 00
indeterminate — keep working
00 does not mean "undefined" and does not mean 0. It means the fraction is hiding a common factor — go find it.
(x − 2)(x + 2)x − 2 = x + 2
valid everywhere except x = 2 itself
x + 2 → 2 + 2 = 4
now substitution works
Cancelling is legal because a limit never needs the point itself — x gets arbitrarily close to 2 without ever equalling 2, and away from 2 the two expressions are identical.
| Question | Asks | Here |
|---|---|---|
| f(2) | where IS the function at 2? | undefined — the hole |
| limit as x → 2 | where is it HEADING near 2? | 4 |
Find the limits of x³ − 8x − 2 as x → 2, and of √x + 5 − 3x − 4 as x → 4.
x³ − 8 = (x − 2)(x² + 2x + 4)
difference of cubes: a³ − b³ = (a − b)(a² + ab + b²)
x² + 2x + 4 → 4 + 4 + 4 = 12
cancel, then plug in
× (√x + 5 + 3) top and bottom
a root on top: use the conjugate
top: (x + 5) − 9 = x − 4
(a − b)(a + b) = a² − b² kills the root
1√x + 5 + 3 → 13 + 3 = 16
the x − 4 cancels
After multiplying, do not expand the denominator — leave it as (x − 4)(√x + 5 + 3) so the x − 4 can cancel. Answering 6 instead of 16 means the fraction was turned upside down.
A root on the bottom instead: the limit of x − 16√x − 4 as x → 16. Read x − 16 as a difference of squares, (√x − 4)(√x + 4). The √x − 4 cancels, leaving √x + 4 → 4 + 4 = 8.
Not 00 but 30. Find the limit of x + 2x − 1 as x → 1.
1 + 21 − 1 = 30
plug in first, as always
x = 0.99: 2.99−0.01 = −299
from the left: plunging
x = 1.01: 3.010.01 = 301
from the right: soaring
the limit does not exist
a vertical asymptote at x = 1
What is different: there is no common factor to cancel. A nonzero number over 0 never settles on a value.
Only 00 sends us looking for a factor. Any other number over 0 means the function blows up, so the limit does not exist.
A negative target. Find the limit of x² + x − 6x² + 3x as x → −3.
9 − 3 − 69 − 9 = 00
(−3)² = 9, with brackets
(x + 3)(x − 2) / [x(x + 3)] = x − 2x
x + 3 is the hidden factor
−3 − 2−3 = −5/−3 = 53
two negatives make a positive
What is different: the factor is x + 3, not x − 3. The hidden factor always has the target as its zero.
Two sides that disagree. f(x) = x + 4 for x < 1 and f(x) = 3x for x ≥ 1. Find the limit as x → 1.
from the left: 1 + 4 = 5
the x < 1 piece
from the right: 3(1) = 3
the x ≥ 1 piece
5 ≠ 3 → does not exist
even though f(1) = 3 exists
What is different: no fraction at all. A limit exists only when the left and right approaches agree.
(a) Find the limit of 1x − 13x − 3 as x → 3. (b) The limit of x² + kx − 3x − 3 as x → 3 exists. Find k and the limit.
(a) 1x − 13 = 3 − x3x
one fraction on top first
3 − x3x ÷ (x − 3) = −13x
3 − x = −(x − 3), so they cancel to −1
→ −13 · 3 = −19
plug in
(b) 9 + 3k − 3 = 0 → k = −2
bottom → 0, so the top must → 0 too
x² − 2x − 3x − 3 = x + 1 → 4
factor (x − 3)(x + 1) and cancel
In (a), losing the minus sign from 3 − x gives +19. In (b), read what is asked — k = −2 and the limit 4 will both be among the choices.
Plug in first: half the items are plain substitution. If you get 00, match the shape — factor a quadratic or cube, use the conjugate for a root, combine a fraction-in-a-fraction. The 10-second check: put x = 2.01 (or a value close to the target) into the original and see if it lands near your answer.
The problem
limx→3 x² − 9x − 3
Substituting 3 gives 00, so factor: (x − 3)x + 3x − 3 = x + 3, and x + 3 → 6. Same shape as the worked example — the graph is the line y = x + 3 with a hole at (3, 6), and the limit reads the height of the hole.
