1. · · · T HE P RECISE D EFINITION OF A L IMIT · · ·
(SMS1102 C ALCULUS 1)
L05-M041
http://staff.iium.edu.my/suryadi/
DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 1 / 15

2. T HE P RECISE D EFINITION OF A L IMIT I NTRODUCTION
”arbitraly” limit
”gets arbitrarily close to”
⇓
⇓
”precisely defined (we chose)”
”precise limit”
DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 2 / 15

3. T HE P RECISE D EFINITION OF A L IMIT I NTRODUCTION
EXAMPLE 1
Consider the function
y = 2x − 1 (1)
near x0 = 4. Intuitively it is clear that y is
close to 7 when x is close to 4, so
lim (2x − 1) = 7 (2)
x→4
However, how close to x = 4 does x have
to be so that y = 2x − 1 differs from 7 by,
say, less than 2 units?
SOLUTION 1
For what value of x is |y − 7| < 2?. In term
of x F IGURE : Keeping x within 1 unit
or x0 = 4 will keep y within 2 units
|y − 7| = |(2x − 1) − 7| = |2x − 8| (3)
of y = 7
|2x − 8| < 2 (4)
−1 < x − 4 < 1 (5)
DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 3 / 15

4. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT
To define δ > 0 so that keeping x within
the interval (x − δ, x + δ) will keep f (x)
1 1
within the interval L − ,L +
10 10
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5. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT
D EFINITION
Let f (x) be defined on an open interval about x0 ,
except possibly at x0 itself. We say that the limit of
f (x) as x approaches x0 is the number L, and
write
lim f (x) = L, (6)
x→x0
if, for every number > 0, there exists a
corresponding number δ > 0 such that for all x,
0 < |x − x0 | < δ =⇒ |f (x) − L| < . (7)
N OTE :
1 When the interval of values δ about x0 is
symmetric, we could take δ to be half the
length of that interval.
2 When such symmetry is absent, we can take
δ to be the distance from x0 to theinterval’s
nearer endpoint.
DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 5 / 15

6. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT
the limit of f (x) as x approaches x0 is the number L,
lim f (x) = L
x→x0
if, for every number > 0, there exists a corresponding number δ > 0
such that for all x,
0 < |x − x0 | < δ =⇒ |f (x) − L| < .
DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 6 / 15

7. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT
the limit of f (x) as x approaches x0 is the number L,
lim f (x) = L
x→x0
if, for every number > 0, there exists a corresponding number δ > 0
such that for all x,
0 < |x − x0 | < δ =⇒ |f (x) − L| < .
DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 7 / 15

8. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT
the limit of f (x) as x approaches x0 is the number L,
lim f (x) = L
x→x0
if, for every number > 0, there exists a corresponding number δ > 0
such that for all x,
0 < |x − x0 | < δ =⇒ |f (x) − L| < .
DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 8 / 15

9. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT
EXAMPLE 2
Show that lim (5x − 3) = 2
x→1
SOLUTION
In definition of limit: x0 = 1, f (x) = 5x − 3
and L = 2, then for ∀ > 0, ∃δ > 0 such
that ∀x
0 < |x − 1| < δ =⇒ |f (x) − 2| < (8)
Fine δ from the −inequality,
|(5x − 3) − 2| = |5x − 5| < (9)
5|x − 1| < (10)
|x − 1| < /5 (11) F IGURE : If f (x) = 5x − 3, then
0 < |x − 1| < /5, guaranties that
Take δ = /5. Any smaller positive δ will |f (x) − 2| <
make 0 < |x − 1| < δ ⇒ |f (x) − 2| <
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10. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT
EXAMPLE 3
Prove the following limits:
lim x = x0 (12) lim k = k (13)
x→x0 x→x0
DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 10 / 15

11. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT
E XERCISE 2.3
Use the graph to find a δ > 0 such that for all x, 0 < |x − x0 | < δ =⇒ |f (x) − L| < .
DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 11 / 15

12. T HE P RECISE D EFINITION OF A L IMIT D EFINITION OF L IMIT
E XERCISE 2.3
Use the graph to find a δ > 0 such that for all x, 0 < |x − x0 | < δ =⇒ |f (x) − L| < .
DR. SU (CTS-KOS•IIUM) LIMIT & CONTINUITY F EBRUARY 27, 2012 12 / 15

13. T HE P RECISE D EFINITION OF A L IMIT F INDING δ A LGEBRAICALLY FOR G IVEN
H OW TO FIND A LGEBRAICALLY A δ FOR A G IVEN f , L, x0 , AND
The process to finding δ > 0 such that for all x
0 < |x − x0 | < δ =⇒ |f (x) − L| < .
can be accomplished in two steps:
1 Solve the inequality |f (x) − L| < to find an open interval (a, b) containing x0 on
which the inequality holds for all x = X0
2 Find a value of δ > 0 that places the open interval (x0 − δ, x0 + δ) centered at x0
inside the interval (a, b). The inequality |f (x) − L| < will hold for all x = x0 in
this δ-interval.
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14. T HE P RECISE D EFINITION OF A L IMIT F INDING δ A LGEBRAICALLY FOR G IVEN
E XAMPLE 4
For the limit
√
lim x −1=2 (14)
x→5
find a δ > 0 that work for = 1. That is, find a δ > 0 such that for all x,
√
0 < |x − 5| < δ =⇒ | x − 1 − 2| < 1.
S OLITION
1 Solve the inequality
√
| x − 1 − 2| < 1 to find an interval
containing x0 = 5 on which the
inequality holds for al x = x0
2 Find a value δ > 0 of to place the
centered interval 5 − δ < x < 5 + δ
(centered at x0 = 5 ) inside the
interval (2, 10).
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15. T HE P RECISE D EFINITION OF A L IMIT F INDING δ A LGEBRAICALLY FOR G IVEN
E XAMPLE 5
Prove that limx→2 = f (x) = 4
x 2, x = 2
f (x) = (15)
1, x = 2
S OLITION
To show that given > 0 there is exist δ > 0 such
that for all x, 0 < |x − 2| < δ ⇒ |f (x) − 4| < .
1 Solve the inequality |f (x) − 4| < to find an
interval containing x0 = 2 on which the
inequality holds for al x = x0
2 Find a value δ > 0 of to place the centered
−
interval 2√ δ < x√ 2 + δ inside the
<
interval 4− , 4+ .
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