SME
Calculus I · Problem type
Find δ for a given ε (a line)
Use this for “For lim (3x − 1) = 5 as x → 2, find δ when ε = 0.06.”
Chart
Within δ = 0.02 of x = 2, f stays within ε = 0.06 of 5 (f(1.98) = 4.94, f(2.02) = 5.06)
This is an example. Type your own values to start a new problem.
lim
x
→
(
m
x
+
(
b
)
)
=
L
δ
=
ε
|
|
=
Formulas
L
=
m
×
a
+
b
5 = 3 × 2 + (−1)
δ
=
ε
÷
|
m
|
0.02 = 0.06 ÷ |3|
Assumptions
•
lim f(x) = L as x → a means: for every ε > 0 there is a δ > 0 with |f(x) − L| < ε whenever 0 < |x − a| < δ.
•
For a line f(x) = mx + b, |f(x) − L| = |m| × |x − a|, so δ = ε ÷ |m| works (any smaller δ does too).
•
When ε ÷ |m| does not end, the page rounds δ; on a proof, round δ down, since a smaller δ still works.
Step-by-step
WE KNOW
m = 3, b = −1, a = 2, ε = 0.06
FIND
the limit (L), allowed distance from a (δ)
Step 1 · Find the limit (L)
Use L = m × a + b
A
line
is
continuous,
so
its
limit
at
a
is
its
value
there.
L
=
3
×
2
+
(−1)
L
=
5
Step 2 · Find allowed distance from a (δ)
Use δ = ε ÷ |m|
|
f
(x)
−
L
|
=
|
m
|
×
|x
−
a
|
<
ε
when
|x
−
a
|
<
ε
÷
|
m
|.
δ
=
0.06
÷
|3|
δ
=
0.02
Check
Put the numbers back in. Both sides should match.
5 = 3 × 2 + (−1) ✓
0.02 = 0.06 ÷ |3| ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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