1.5
Limits as
→ and Continuity
Definition 8 We said a function has limit at point if the following equation is satisfy lim → () = (1.1) or lim →+ () = lim→− () =
We said the function have no limit at point if above equation or definition does not holds. In general, we have three type
(a) lim
→+ ()6= lim→− () ;
Example 20 () = ||
and = 0 (b) lim
→+ () or lim→− () does not exist;
Example 21 () = 1
2 and = 0
(c) Oscillating
Example 22 () = sin 1
and = 0 see FIG 1
5 2.5 0 -2.5 -5 1 0.75 0.5 0.25 0 -0.25 -0.5 -0.75 -1 x y x y FIG 1 sin1 Properties of limits
Theorem 6 Let ∈ R, ∈ N, (1) lim → = ; (2) lim → = ; (3) lim → =
Theorem 7 Let ∈ R, ∈ N, let be functions with the following limits lim → () = lim→ () = Then (1) lim →( ()) = ; (2) lim →( ()± ()) = ± ; (3) lim → () () = ; (4) lim → () () = provided 6= 0; (5) lim →[ ()] = ; (6) lim → p () = √ provided lim
→ () is not negative if is even.
How to find lim
→ ()?
Use the Theorem 6, 7 we have following formulas A:
Theorem 8 If is a polynomial function and ∈ R. Then lim → () = () Example 23 lim →5( 2 − 7 + 4) = −6
Theorem 9 If is rational functions given by () = ()
() Then lim → () = () = () () provided () 6= 0 Example 24 lim →4 2+ 1 5 + 2 = 42+ 2 5× 4 + 2 = 17 22
Theorem 10 Let ∈ N The following limit is valid for all if is odd and is valid for 0 if is even
lim
→
√
= lim 1 = √;
Theorem 11 If are functions such that lim
→ () = lim→ () = () then lim → ( ()) = () ; Example 25 If lim →0( 2+ 4) = 02+ 4 = 4 and lim →4 √ = 4 then lim →0 √ 2+ 4 = 2 B:(0 0 model) Let () = ()
() be rational function, if lim→ () = 0 and lim→ () = 0
Then we must be use following Theorem and tools to find lim
→ () = lim→
() () Theorem 12 Let ∈ R, let = ( ) be a open interval contain and let
() = () ∀ ∈ 6= If lim → () = then lim → () = lim→ () = Example 26 For () = 2 − 7 + 6 2− 36 µ 0 0 as → 6 ¶ since 8 6 4 2 0 1 0.5 0 -0.5 -1 x y x y
2 − 7 + 6 2− 36 = (− 6) ( − 1) (− 6) ( + 6) = − 1 + 6 = () for 6= 6 and lim →6 () = lim→6 − 1 + 6 = 5 12 Thus by Theorem 12 we have
lim →1 () = lim→1 () = 3 Example 27 For () = √ + 1− 1 µ 0 0 as → 0 ¶ since 5 3.75 2.5 1.25 0 1 0.875 0.75 0.625 0.5 0.375 x y x y √ + 1− 1 = √ + 1− 1 √ + 1 + 1 √ + 1 + 1 = √ 1 + 1 + 1 Thus by Theorem 12 we have
lim →0 √ + 1− 1 = lim→0 1 √ + 1 + 1 = 1 2 C: The Squeeze Theorem
Theorem 13 If () ≤ () ≤ () for all ∈ 6= and if lim → () = = lim→ () then lim → () =
How to use the Squeeze Theorem
(1) Use −1 ≤ sin ≤ 1 −1 ≤ cos ≤ 1 ∀ and Example 28 Find lim
→0|| sin 1 since −1 ≤ sin ≤ 1 =⇒ − || ≤ || sin ≤ || and lim →0− || = 0 and lim→0||
Thus by the squeeze Theorem we have lim →0|| sin 1 = 0 (2) Dominate-test If | ()| ≤ () and lim → () = 0 then lim → () = 0
Example 29 Find lim
→0 sin 1 since ¯ ¯ ¯ ¯ sin1 ¯ ¯ ¯ ¯ = || ¯ ¯ ¯ ¯sin1 ¯ ¯ ¯ ¯ ≤ || and lim →0|| = 0
Thus by Dominate-test we have
Continuity:
Definition 9 (Continuity at a point) A function is continuous at = if the following conditions are satisfy
(1) () is defined; (2) lim
→ () exists ( lim→+ () = lim→− ();
(3) lim
→ () = ()
Another we say function is not continuous(discontinuous) at point , if one of following three conditions is hold.
(1)(Removable discontinuity) The is not defend at = ( defined in ( ) expect ;
(2)(Nonremovable discontinuity) The limit of () does not exist at = µ
lim
→+ ()6= lim→− ()
¶ ;
(3)(Removable discontinuity) The limit of () exists at = but ()6= lim
→ ()
Definition 10 Continuity on an Open interval: is continuous on ( ) if it is continuous at each point in ( )
Example 30 Let () = ½ 9− 2 , if 0 ≤ 2 052+ 3 if 2≤ ≤ 4 Determine whether () is continuous at = 2
One sided limits and Continuity on a Closed interval [ ] A:one sided limits ( lim
→+ () lim→− ())
Example 31 lim
→0+
√ = 0
Example 32 Find the limit of () = √4− 2 as approaches −2 from
the right, i.e
lim
→−2+ () = 0
Why, since the domain of is [−2 2].
B:The Greatest integer function (Gauss function) () = []
1 0 xy xy Example 33 lim →0−[] =−1 lim→0+[] = 0
Example 34 Find lim
→1 £ (− 1)2¤ 2− 1 since lim →1− £ (− 1)2¤ 2− 1 = lim→0− 0 2− 1 = 0 lim →1+ £ (− 1)2¤ 2− 1 = lim→0+ 0 2− 10 Thus lim →1 () = £ (− 1)2¤ 2− 1
Exercise 1 Find lim
→1(1− + [] + [1 − ]) =?
Example 35 Find lim →0− || and lim→0+ ||
Theorem 14 (The existence of limit) The limit of () as approaches is if and only if (⇔)
lim
→+ () = lim→− () =
Definition 11 (Continuity on a closed interval [ ]) We say is continu-ous on [ ], if
(1) is continuous on ( ) and (2) lim
→ () = () (continuous from right at ) and lim→ () = ()
(con-tinuous from left at ).
Example 36 () = √1− 2 the domain of is [−1 1] is continuous
on (−1 1) and lim →−1+ √ 1− 2 = 0 lim →1− √ 1− 2 = 0 Thus is continuous on [ ] Properties of Continuity
Theorem 15 If ∈ R and are continuous at = then the following functions are also continuous at
(1) Scalar multiple:
(2) Sum and difference ± (3) Product:
(4) Quotient:
if () 6= 0
Example 37 (1) Polynomial functions () = +−1−1+· · ·+1+
0
(2) Rational function: () = ()() ()6= 0 (3) Radical function: () = √
Theorem 16 If is continuous at and is continuous at () then the composite function ◦ () = ( ()) is continuous at
Theorem 17 (Intermediate Value Theorem) If is continuous on [ ] and is any number between () and () then ∃ ∈ [ ] such that () = Remark 1 Continuity is necessarily condition.
Example 38 5 2.5 0 -2.5 -5 0.5 0 -0.5 x y x y
Example 39 (Application) Use the Intermediate Value Theorem to show that the
() = 3+ 2− 1 has a zero in [0 1]
Proof. Since is continuous on [0 1] and (0) = 03+ 2
· 0 − 1 = −1 (1) = 13+ 2
· 1 − 1 = 2
Moreover (0) = −1 0 2 = (1) Thus by Intermediate Value Theorem ∃ ∈ [0 1] such that () = 0
