SME
Calculus I
Limits and continuity
Use this for “Find the limit of (x² + x − 6) ÷ (x − 2) as x → 2.”
Chart
Slant asymptote y = x + 3
Hole at (2, 5)
f(1.9) = 4.9
As x → 2 from the left, f(x) → 5; from the right, f(x) → 5.
So the limit is 5, though f(2) has no value.
This is an example. Type your own values to start a new problem.
lim
x
→
x²
+
x
+
(
)
x
−
=
L
f(
x
)
=
f(x)
N
Top at x = a
worked out (not typed)
Formulas
N
=
a
2
+
a
×
b
+
c
0
=
2
2
+
2
×
1
+
(−6)
L
=
2
×
a
+
b
5 = 2 × 2 + 1
f
(
x
)
=
x
2
+
b
×
x
+
c
x
−
a
4.9
=
1.9
2
+ 1 × 1.9 + (−6)
1.9 − 2
Assumptions
•
f(x) = (x² + bx + c) ÷ (x − a). The limit is the value f(x) approaches as x → a, not f(a).
•
Top 0 at a (0/0): x − a is a factor of the top. Cancel it (allowed, since x ≠ a) and put in x = a.
•
Top k ≠ 0 at a (k/0): f grows without bound near a, a vertical asymptote; there is no finite limit. The sign of k over the sign of x − a gives +∞ or −∞ on each side.
Step-by-step
WE KNOW
a = 2, b = 1, c = −6, x = 1.9
FIND
top at x = a (N), the limit (L), f(x) there (f(x))
Step 1 · Find top at x = a (N)
Use N = a² + a × b + c
Put
x
=
a
into
the
top.
If
it
is
0,
x
−
a
is
a
factor;
if
not,
the
quotient
is
k
0
.
N
=
2
2
+
2
×
1
+
(−6)
N
=
0
Step 2 · Find the limit (L)
Use L = 2 × a + b
Factor:
x
2
+
bx
+
c
=
(
x
−
a
)(
x
+
a
+
b
).
Cancel
x
−
a
:
f
(
x
)
=
x
+
a
+
b
for
x
≠
a
,
which
nears
2
a
+
b
as
x
→
a
.
N
=
0:
the
top
is
0
(
0
0
),
so
factor
and
cancel
L
=
2
×
2
+
1
x
=
1.9:
f
(
x
)
=
1.9
+
2
+
1
=
4.9
x
=
1.99:
f
(
x
)
=
1.99
+
2
+
1
=
4.99
x
=
2.01:
f
(
x
)
=
2.01
+
2
+
1
=
5.01
L
=
5
Step 3 · Find f(x) there (f(x))
Use f(x) = (x² + b × x + c) ÷ (x − a)
Put
x
into
the
top
and
the
bottom,
then
divide.
f
(
x
)
=
1.9
2
+ 1 × 1.9 + (−6)
1.9 − 2
f
(
x
)
=
4.9
Check
Put the numbers back in. Both sides should match.
0
=
2
2
+
2
×
1
+
(−6)
✓
5 = 2 × 2 + 1 ✓
4.9
=
1.9
2
+ 1 × 1.9 + (−6)
1.9 − 2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Make a piecewise function continuous
Use this for “Find k so that f(x) = 3x + k (x < 2), x² + 1 (x ≥ 2) is continuous.”
The intermediate value theorem
Use this for “Show that x³ + x − 1 = 0 has a root between 0 and 1.”
The limit of sin(kx) ÷ (mx)
Use this for “Find the limit of sin(3x) ÷ (2x) as x → 0” or “Find k so that sin(kx) ÷ (2x) → 3 as x → 0.”
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.”
Limits at infinity of a quotient
Use this for “Find the limit of (5x² − x) ÷ (2x² + 7) as x → ∞.”
Kinds of discontinuity
Use this for “Is the discontinuity of (x² − 4) ÷ (x − 2) at x = 2 removable?”