Teaching Demonstration · August 18, 2026
Jesica M. Anoche
Department of Mathematics and Statistics
When we say
“L is the limit of f(x), as x approaches a”
we mean
f(x) → L as x → a
Read it as motion: push x toward a and watch where f(x) is dragged.
| x | 1 | 0.5 | 0.25 | 0.10 | 0.01 | 0.001 | 0.0001 |
|---|---|---|---|---|---|---|---|
| f(x) | 4 | 2.5 | 1.75 | 1.3 | 1.03 | 1.003 | 1.0003 |
| x | −1 | −0.5 | −0.25 | −0.10 | −0.01 | −0.001 | −0.0001 |
|---|---|---|---|---|---|---|---|
| f(x) | −2 | −0.5 | 0.25 | 0.7 | 0.97 | 0.997 | 0.9997 |
The values of f(x) approach 1 as x approaches 0. In symbols,
Definition 1.2 Let f be a function which is defined for all x on the open interval I containing a, except possibly at a itself. The limit of f(x) as x approaches to a is L, written
limx→af(x) = L
if for every ε > 0, however small, there exists a δ > 0 such that
|f(x) − L| < εwhenever0 < |x − a| < δ
The gold band represents the allowed vertical distance around L, while the blue band represents the corresponding horizontal distance around a.
If c is a constant, then for any real number a,
limx→ac = c
If b and c are constants, then
limx→a(bx + c) = ba + c
For any real number a,
limx→ax = a
If limx→af(x) = L and limx→ag(x) = M, then
limx→a[f(x) + g(x)]=limx→af(x)+limx→ag(x)=L + M
If limx→af1(x) = L1 and limx→af2(x) = L2, then
limx→a[f1(x) · f2(x)]=limx→af1(x)·limx→af2(x)=L1 · L2
If limx→af(x) = L and n is any positive integer, then
limx→a[f(x)]n=[limx→af(x)]n=Ln
If limx→af(x) = L, then
limx→an√f(x)=n√limx→af(x)=n√L
provided n√L ∈ ℝ
If limx→af(x) = L and limx→ag(x) = M, M ≠ 0, then
limx→af(x)g(x)=limx→af(x)limx→ag(x)=LM
Theorem 9 If F and G are two functions such that F(x) = G(x) for all x ≠ a, and if limx→aG(x) exists, then limx→aF(x) exists, and the two are equal.
Try it Substituting x = 2 gives 00 — meaningless. Theorem 8 does not apply.
Look closer For every x ≠ 2,
F(x) = x2 − 2xx − 2 = x(x − 2)x − 2 = x = G(x).
Apply it By Theorem 9, since F(x) = G(x) for every x ≠ 2,
limx→2x2 − 2xx − 2 = limx→2x = 2
One point is punched out of the line. The limit does not care — it only watches the approach.
Try it Again 00. Factoring will not help here — there is a root in the way.
The move Multiply above and below by the conjugate √x+1 + 1.
√x+1 − 1x·√x+1 + 1√x+1 + 1=xx(√x+1 + 1)
Cancel For every x ≠ 0 the x divides out, so F(x) = 1√x+1 + 1.
limx→01√x+1 + 1=1limx→0(√x+1 + 1)=11 + 1=½
The graph has a hole at x = 0, yet both sides close in on one half.
Theorem 10. The limit limx→af(x) exists and equals L if and only if both one-sided limits exist and are equal:
limx→a−f(x) = L and limx→a+f(x) = L.
Example Evaluate limx→0|x|x.
Left side If x < 0, then |x| = −x, so limx→0−|x|x = limx→0−−xx = −1.
Right side If x > 0, then |x| = x, so limx→0+|x|x = limx→0+xx = 1.
limx→0−|x|x = −1 ≠ 1 = limx→0+|x|x, so limx→0|x|x does not exist.
Left-hand limit: −1. Right-hand limit: 1. Since they disagree, the two-sided limit does not exist.
limx→4x2 − 16x − 4
limx→4x2 − 16x − 4 = limx→4(x + 4)(x − 4)x − 4 = limx→4(x + 4) = 4 + 4 = 8
Thank you for listening
and God bless.
Jesica M. Anoche · August 18, 2026