ChapterOne Exam Review
DN
*
*
O When does
Limits
a limit not exist rule of 3
I F O
毒
= =
T
. ,
*
·
FCX1 approaches a different # from the right-left of C
.
muu〜 Solving limits Algebraically In direct substitution ·
F(X) increases or decreases without bound as X-* C
Always begin with direct substitution if 8-1 F . . ·
F(x) Occilates between two fixed values as X-* C
.
XmF(X
look for common denominators if complex fraction Special notes 3
=
·
3-
〜
·
Simplify or Cancel (-X/Xis-1) You can have undefined values and limit still exists !
·
Try to rationalize numerator or denominator if applicable If you know a point exF(4) = 2
,
that tells you nothing
Infinite Limits(1 .S)- about the limit ,
and Vice versa if you know limit you don't
DetermineWhether f(x) approaches +oo or-oo as X-* # From nesasarly know the -value at that X value
the left-right there ,
is a vertical asymptote and you need to
plug
If you see absolute value make a piecewise tright
",
eximthen 뼉
X 3#
in values left-right of number and determine + piug in vaies closeto it toe oet --left
(x[t ℃ ; ×《 H
-
left hand+ right hand limits
Continuity Removable Discontinuity Hole canceorfromtopbottom =
- you can
ess
NOn-Removeable Dis Continuity Vertical asymptote =
DiscontinuitiesWith trig Functions =
Trig function Periods =
Limits w/
thig functions
m 〜
VAI
How to find- (may ask for
Tan (Ot =+ all other GT
If continuous everywhere just plug
removable -
Discontinuity ·
in and get your value
1) Set B equal to trig functions asymptote nig
function asymptotes = ·
Reciprical/one-sided limits =
Bvalve
Lif reciprocal check where It is undefined it will either be
X 0
-
y = 3X = 0 (- Words (beside cos) X # OO orDNE
=
= (S(3X eX = OTI IT , 2TT
, Lleft/right handed plug in value close to # then
2)Solve for X-- first value for answer all Others(nosin) X = , ↳ If not one-sided
DNE
3) then get period (either TT) ↑memorize
GTT or :
* Left-Right handtrig limitso
4) divide period by B-valve cottan periods :
点:
5) then divide that by normal period (2011) Sin-cos
&
SectCS + OTT
Answer non-removeable :
discontinuity at o CSC + (Ot VAS at Ott TT 2Tt
*
, ,
+ K Where I is & secttan vas St it at
integer
+
,
an .
use properties
I VI Other + IVT-
it will ask to prove a zero or passes
through a certain valve
Of lims when
Greatest Integer
doing composite
Since FCX1 is Continuous on [a b] and Fal =_ and F(bl
,
=
m
due to the intermediate value theorem there
F(X) [X1] =
or 11X11 non-continous
function One number C Such that F(c) Q Or
=
given value
must be at least
limS YmoF(g(X11 =
F(xiMog(x)
L
greatest integer less than or earal to in put
f(2) [121] =
= 2
F[ -
031: 11 - . 31に - 1
0
Any number divided by 00 is zero : =
0
FC211 :
[1211 ]: 2 F(2 71 /1-2 711
.
: .
=. 3 SUM OF (UbeS -
Any number divided by a super small number is = 00
F
(2.71 f(-3 1) 11-3 :
(12 71 .
: 2 .
= .
111 = -
4
(P b3+ =
(Afb) (@2Fab b2) + 증 = 00
즈
. 0001
0
-00
=
001
-
you may get asked lm[IX1]
CP b3 - =
(aGb) (@0ab + b)
you ms break up into left + right
西
Pieces
DN
*
*
O When does
Limits
a limit not exist rule of 3
I F O
毒
= =
T
. ,
*
·
FCX1 approaches a different # from the right-left of C
.
muu〜 Solving limits Algebraically In direct substitution ·
F(X) increases or decreases without bound as X-* C
Always begin with direct substitution if 8-1 F . . ·
F(x) Occilates between two fixed values as X-* C
.
XmF(X
look for common denominators if complex fraction Special notes 3
=
·
3-
〜
·
Simplify or Cancel (-X/Xis-1) You can have undefined values and limit still exists !
·
Try to rationalize numerator or denominator if applicable If you know a point exF(4) = 2
,
that tells you nothing
Infinite Limits(1 .S)- about the limit ,
and Vice versa if you know limit you don't
DetermineWhether f(x) approaches +oo or-oo as X-* # From nesasarly know the -value at that X value
the left-right there ,
is a vertical asymptote and you need to
plug
If you see absolute value make a piecewise tright
",
eximthen 뼉
X 3#
in values left-right of number and determine + piug in vaies closeto it toe oet --left
(x[t ℃ ; ×《 H
-
left hand+ right hand limits
Continuity Removable Discontinuity Hole canceorfromtopbottom =
- you can
ess
NOn-Removeable Dis Continuity Vertical asymptote =
DiscontinuitiesWith trig Functions =
Trig function Periods =
Limits w/
thig functions
m 〜
VAI
How to find- (may ask for
Tan (Ot =+ all other GT
If continuous everywhere just plug
removable -
Discontinuity ·
in and get your value
1) Set B equal to trig functions asymptote nig
function asymptotes = ·
Reciprical/one-sided limits =
Bvalve
Lif reciprocal check where It is undefined it will either be
X 0
-
y = 3X = 0 (- Words (beside cos) X # OO orDNE
=
= (S(3X eX = OTI IT , 2TT
, Lleft/right handed plug in value close to # then
2)Solve for X-- first value for answer all Others(nosin) X = , ↳ If not one-sided
DNE
3) then get period (either TT) ↑memorize
GTT or :
* Left-Right handtrig limitso
4) divide period by B-valve cottan periods :
点:
5) then divide that by normal period (2011) Sin-cos
&
SectCS + OTT
Answer non-removeable :
discontinuity at o CSC + (Ot VAS at Ott TT 2Tt
*
, ,
+ K Where I is & secttan vas St it at
integer
+
,
an .
use properties
I VI Other + IVT-
it will ask to prove a zero or passes
through a certain valve
Of lims when
Greatest Integer
doing composite
Since FCX1 is Continuous on [a b] and Fal =_ and F(bl
,
=
m
due to the intermediate value theorem there
F(X) [X1] =
or 11X11 non-continous
function One number C Such that F(c) Q Or
=
given value
must be at least
limS YmoF(g(X11 =
F(xiMog(x)
L
greatest integer less than or earal to in put
f(2) [121] =
= 2
F[ -
031: 11 - . 31に - 1
0
Any number divided by 00 is zero : =
0
FC211 :
[1211 ]: 2 F(2 71 /1-2 711
.
: .
=. 3 SUM OF (UbeS -
Any number divided by a super small number is = 00
F
(2.71 f(-3 1) 11-3 :
(12 71 .
: 2 .
= .
111 = -
4
(P b3+ =
(Afb) (@2Fab b2) + 증 = 00
즈
. 0001
0
-00
=
001
-
you may get asked lm[IX1]
CP b3 - =
(aGb) (@0ab + b)
you ms break up into left + right
西
Pieces
