Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
6 Tangent and Velocity(2.1)
Self reading of chapter 2.1
7 Limit of Function(2.2)
Ex 1. Consider function f ( x ) =
x 2 − 3x + 2
. The function f(x) is definitely undefined for
x −1
x=1. Let’s investigate the behavior of f(x) for values near x=1. The following
table shows values f(x) for values of x close to but not equal to 1.
x
f(x)
0.8
-1.2
0.9
-1.1
0.99
-1.01
.999
-1.001
1.001
-0.999
1.01
-0.99
1.1
-0.9
1.2
-0.8
One can see from the table that f(x) getting close to -1 when x approaches 1. For a
simple example like this, it is possible to get the same result algebraically since
x 2 − 3x + 2
=
x −1
( x − 1)( x − 2 )=
x −1
x − 2 , so this is a line with a hole in x=1.
We shall write lim
f ( x ) =lim
x →1
x →1
( x − 1)( x − 2 ) =lim x − 2 =−1
x 2 − 3x + 2
=lim
x →1
x →1
x −1
x −1
f ( x ) = L means: “the limit of f(x) as x approaches a equals L”.
Def: The equation lim
x→a
Note, the function f(x) isn’t required to be defined at x=a. It just has to get closer to L
as x gets closer (but not equal) to a.
f (a)
L+ε
Note: A more rigorous ( ε , δ ) -definition of limits talking
L
about existence of an open interval I a =( a − δ , a + δ ) for any
L−ε
choice of arbitrary small interval I L =( L − ε , L + ε ) (where
both ε , δ > 0 ), such that ∀x ∈ I a imply f ( x ) ∈ I L .
a−δ
a a+δ
0.1667
0.1667
0.1667
0.1667
0.1667
Ex 2. Let f ( x ) =
x +9 −3
f ( x)
, find lim
x →0
x2
2
0.1667
0.1667
0.1666
0.1667
0.1666
0.1666
0.1667
0.1666
0.1667
0.1666
0.1667
The algebraic approach gives
0.1665
-1.5
-1
-0.5
0
0.5
1
1.5
2
0.1665
-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
-4
x 10
x2 + 9 − 3
x2 + 9 − 3 x2 + 9 + 3
1
1
lim
=
lim
= lim
=
2
2
x →0
x →0
x
x
x 2 + 9 + 3 x →0 x 2 + 9 + 3 6
However when one uses a table for really small values of x we may arrive in the
wrong answer (this is problem of computer accuracy, in numerator we get 3.0…0 -3,
so it considered zero). See also the 2 fine scaled graphs of the function above.
0.1
Course: Accelerated Engineering Calculus I
x
f(x)
±5.E-06
0.1667
±1.E-06
0.1665
±5.E-07
0.1670
Instructor: Michael Medvinsky
±1.E-07
0.1776
-5.E-08
0.1776
-1.E-08
0.0000
Some limits are easy to get from table or sketch like even non
polynomial ones like lim
x →0
5.E-09
0.0000
sin(x)/x
1.0001
1
sin x
= 1 (see the image), but some
x
0.9999
0.9998
0.9997
may be tricky and some not exists, and some may mislead
like in following example.
π
x
1.E-09
0.0000
0.9996
0.9995
0.9994
0.9993
-0.05
-0.04
-0.03
-0.02
-0.01
0
0.01
0.02
0.03
0.04
1
n
sin , one get that for x = for any integer n>0,
Ex 3. If we create a table to find lim
x →0
one gets sin nπ = 0 which can mislead to the WRONG guess
sin
that lim
x →0
π
x
= 0 . This is wrong because for example for
1
0.8
wrong
0.6
2
π
2
one gets sin (10n + 1) =
one
x= n
1 and for x =
10 + 1
2
3 (10n + 1)
3π
π
−1 . Actually the function sin is highly
gets sin (10 + 1) =
2
x
n
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-0.1
-0.08
-0.06
-0.04
0.02
0
-0.02
0.04
0.06
0.08
oscillating near x=0 and therefore the limit doesn’t exist.
Def: One Sided Limits: The equation lim f ( x ) = L means: “the limit of f(x) as x
x→a+
approaches a from the right equals L”. Similarly lim f ( x ) = L means: “the limit of
x→a−
f(x) as x approaches a from the left equals L”.
0
1
Ex 4. For a Heaviside function f ( x ) = 
x<0
the limit
x≥0
lim f ( x ) isn’t exists, because of the jump. However lim− f ( x ) = 0
x →0
x →0
and lim f ( x ) = 1 . This gives a hint for the following theorem.
1
0.8
0.6
0.4
0.2
0
-0.2
-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
0.1
x → 0+
f ( x ) is exists if and only if one sided limits exists and equal, i.e.
Theorem: A limit lim
x→a
lim f ( x )= L= lim+ f ( x ) .
x→a−
x→a
Ex 5. In the f(x) func on the graph lim f ( x ) = 0.5 ≠ 2 = lim f ( x ) ,
−
x →1
f ( x ) not exists (there is a jump), but
therefore lim
x →1
lim f ( x ) =
4=
lim+ f ( x ) ⇒ lim f ( x ) =
4.
x → 2−
x→2
x→2
+
x →1
4
2
-0.5
1 2
0.1
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
1
x
Ex
6. lim c c=
;
lim x a; =
lim x 2 a 2 ; =
lim
=
x→a
x→a
x→a
x→a
1
a
Ex 7. Let g (=
x) f ( x + c) . Show that if lim f ( x) = L then lim g ( x) = L .
x →c
x →0
g=
( x) lim f ( x =
+ c)
Solution: lim
x →0
x →0
f=
( y) L
lim
x + c = y →c
7.1 Limit Laws(2.3)
f ( x ) , lim g ( x ) exists and finite. Then
Def: Let c be a constant and lim
x→a
x→a
1) lim  f ( x ) ± g ( x ) = lim f ( x ) ± lim g ( x )
x→a
x→a
x→a
2) lim cf ( x ) = c lim f ( x )
x→a
x→a
3) lim f ( x ) g ( x ) = lim f ( x ) lim g ( x )
x→a
4) lim
x→a
x→a
lim f ( x )
f ( x)
= x→a
g ( x ) ≠ 0 lim g ( x )
g ( x ) xlim
→a
x→a
x→a
( )
( )
2
n
=
xn =
an
lim x lim x n −1 =
lim x lim x lim x n − 2 =
lim x lim x n − 2 =
lim x =
Ex 8. n ∈  + , lim
x→a
x→a x→a
x →a x →a x →a
x →a
x →a
x→a
LL 3
LL 3
(
)
n
Ex 9. Similarly n ∈  + , lim
lim f ( x )
f n ( x) =
x→a
x→a
LL 3
3
3
3=
27 , see it on graph of x3 .
x = 3 and lim x=
Ex 10. Evidence lim
x →3
x →3
(
Ex 11. Another evidence lim sin 2 x =
x →3π /2
(
n
n lim n x
Ex 12. n ∈  + , lim
x=
x→a
x→a
)
n
x →3π /2
n lim
=
x→a
LL 3
Ex 13. Similarly
n ∈  + , lim n f ( x )
=
x→a
n
)
2
( −1)
lim sin x =
( x)
n
n
2
n
n lim x =
=
a , for even n assume a > 0
lim f ( x ) ,
x→a
x→a
for even n assume lim f ( x ) > 0
x →a
Def: Direct Substitution Property: if f(x) “have no problem at a” then
lim f ( x ) = f ( a ) . For now we consider that roots, polynomial, rational and
x→a
trigonometrical functions “have no problem at a” if defined at a.
Ex 14. lim
x →5
3) ( 5 − 2 )( 5 − 3)
( x − 2 )( x −=
=
x
5
3⋅ 2 6
=
5
5
f ( x ) = lim g ( x )
Def: if f ( x ) = g ( x ) in an open interval x ∈ ( a − b, a ) ∪ ( a, a + b ) , then lim
x→a
x→a
Course: Accelerated Engineering Calculus I
( x − 5)
Ex 15. lim
x →0
2
Instructor: Michael Medvinsky
− 25
x 2 − 10 x + 25 − 25
x 2 − 10 x
=
lim
=
lim
=
lim x − 10 =
−10
x →0
x →0
x →0
x
x
x
( x − 2 )( x − 3) ≠ ( 0 − 2 )( 0 − 3) =
( x − 2 )( x − 3) lim = x − 3 =0 − 3 =−3
"0"
Ex 16. lim
however
lim
( )
x→2
x→2
x→2
x−2
0−2
"0"
x−2
( x − 1=
)( x − 2 ) lim
x 2 − 3x + 2
x −1 1
Ex 17. xlim
lim
=
=
2
2
2
x
x
→2
→
→
x −4
x+2 4
( x − 2 )( x + 2 )
Ex 18. lim
+
x →1
x −1
x −1
x −1
x −1
= lim+
= 1 ≠ −1 = lim−
= lim−
x →1 − ( x − 1)
x →1
x − 1 x →1 x − 1
x −1
x −1
x −1
= lim
Ex 19. lim
x →1
x →1
x −1
1
1
1
1
= lim
= = =
x →1
x + 1 lim x + 1 1 + lim x 2
x −1
x +1
x →1
x →1
(
)(
3 − x 2 x ∈ Q
Ex 20. Let f ( x ) = 
x ∉Q
x
)
f ( x) is exists.
, find c such that lim
x →c
3 − x 2 =3 − c 2 =c =lim x , thus
Solution: Actually we need c such that lim
x →c
x →c
−1 ± 1 + 12 −1 ± 13
3 − c 2 =c ⇒ c 2 + c − 3 =0 ⇒ c1,2 =
=
2
2
f ( x ) ≤ lim g ( x )
Theorem: If f ( x ) ≤ g ( x ) in an open interval x ∈ ( a − b, a + b ) , then lim
x→a
x→a
The Squeeze Theorem (Other names: Sandwich Thrm, two
policemen and a drunk theorem): If f ( x ) ≤ g ( x ) ≤ h ( x ) in an
f ( x )= L= lim g ( x ) , then
open interval x ∈ ( a − b, a + b ) , and lim
x→a
x→a
15
10
5
lim h ( x ) = L .
x→a
0
-1
Ex 21. Let see that xlim
→0
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1
eax − 1
(see
= a . Since at 0 ≤ ax < 1 we have 1 + ax ≤ eax ≤
1 − ax
x
the graph), the squeeze theorem provides (in each part of inequality sub 1 and
divide by x): a ≤
eax − 1 1  1
ax

≤ 
=
− 1
→a
x
x  1 − ax  x (1 − ax )
sin x
Ex 22. Let see why lim
= 1 : Let see the areas of “slice of pie”
x →0
x
x
x
AOB and triangles AOB and COB: =
Sπ AOB =
,
π 12
2π
2
1
sin x
1
tan x
S AOB = ⋅ OB ⋅ h =
, SCOB = ⋅ OB ⋅ h ' =
. Using the
2
2
2
2
C
A h’=tanx
1
O x1
B
h=sinx
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
relationship between sizes of these shapes, one write sin x < x < tan x , next
x
tan x
1
<=

→1 , by squeeze theorem we
sin x sin x cos x x →0
x
sin x
1
1
= 1 , and the last trick is lim
got that lim
= lim
= = 1.
x → 0 sin x
x →0
x
→
0
x
x
x
lim
sin x x →0 sin x
divide all by sinx to get 1 <