This document covers the basics of limits, asymptotes, methods of factorisation, finding a continuity condition, continuities, discontinuities, and piecewise functions. Not only does the document include clear, easy to understand detailed notes, but also a practice test at the end with the answers.
LIMITS
Limits : The value that a function approaches as the input
approaches some value
realityvsexpectation
✓
^
,
Expectation :
limfcx )
✗ → a
in undefined
' "
2
GABBY Reality
. ,
: FCX )
define,
, DNE : does not exist
-3 2 I 2 3
1
-
-
-1
Ex : in the
vicinity of ✗ = -2
,
-2 EXP Lim FCX)= 2
✗ → -2
since there is no •
-3 REAR f- C- 2) =D NET
More ex :
Lim fcx )= DNE him f- ( X ) -_ 2
✗ → I ✗ →2
f- (1) = I f- (2) = 0
✗ = -
I Limfcx) =-3 EXP
✗ → -
I
-
4 %••☒
-
f- C- 1) = -
2 REAL
.
-
2 %Aµ☒
✗= I limfcx ) =
-
I
- Bog ✗ → I
Fatone
2
-
-
,
✗= 2 Iim f- 1×7=0
q••☒ -
-2 ✗ → 2
-
f- (2) = 2
,
-
-4
✗ =3 Iim f- ( x )= DNE
✗ → 3
.
f- (3) = I
✗= 4 limfcx ) =3
✗ → 4
f- (4) = 4
, and
-
+ :
✗ at look at function
→
means to the coming
from the
right
✗ → a- means to look at the function
coming
from the left
one sided limits :
2 limits from left and the
coming the
for the variable ( jump )
right same
Two sided limits t limit
coming from left and the
:
the
for the
right same variable .
Always DNE
Ex :
}
Iim f- 1×1=2
.{i)fq*+ne+w
✗ → 1- one sided limits
.am?.. *zLimfCx
Iim FCX ) =3
-- 2 ✗ → It .
Lim FCX ) = DNE → two sided limits
☒ → I
f- (1) = -
I
. sided limits
,
) exists only if the
✗ →a
-2
one sided limits , lim f- ( X)
I and Lim Flx )
✗ → a-
exist
✗ → at
and
agree .
Iim = Iim =/ im
✗→ a ✗ → at ✗ → a-
1 Iim f- ( X ) = 2
✗ → it
-
man -2 Iim fix)= 3
* → -
zt
-
I ¥8 limfcx ) =3
✗ → of
-
Iim f- Cx) =3
- -2
✗ → 0
-
limflx)= DIVE
✗ → I
f- C- 2) = 2
, DISCONTINUITY
continuity : A function is continuous if limflx )=limflx_- flat
✗ → a- ✗ → at
and all exist
Discontinuity : A function is discontinuous at ✗ =a if exp =/ real
The function has a
discontinuity at X=a if we lift
at ✗ =a to draw the of the function
our
pen graph .
..¥
^
/
}
"" "
:÷÷÷sa+×=
✗ = 10
,
8- since exp = real and we don't
I f- ( 17=2
6- *a→
I
4-
Iim
}
FCX )= DNE
2- neat 1 ✗ → 5 fcx ) is discontinuous at
Iimfcx )= 4 EXP
,
✗ =5 since exp # real and
I 1 I 1 I > ✗ → 5-
we must lift our pen
4 8 f-(5) 6 REAL =
JUMP DISCONTINUITY
limflx )=6
✗ → 5-1
I
} }f¥fi§,ndiI%Ip¥¥a
limfcx )=5 fix ) discontinuous at Iim f- 1×1=+00 at
✗ → 8 =8 since
✗ exp =/ real ✗ → 10 ,
f- (8) = DNE REMOVABLE DISCONTINUITY f- ( lo ) = DNE INFINITE DISCONTINUITY
Types of discontinuity
) ¥ limflx ) but Can
Jump When
linga,ECx they both exist be equal or
: .
✗ → a-
not to f- (a)
none infinite
-
^ are
r n
-
can occur at :
BET breakpoints in piecewise
EA
functions ( 2 > × )
#
>
§
s
absolute values
giving
( ¥¥z )
.
removable :
when limfcx)= limftx ) =/ flat but limfcx ) exists
,,.••none
✗ → a- ✗ → a
exp =/ real s
'
IIMFCX ) #
-
^
same f- (a)
thing as
✗ → a
are man ,+e
s
-
can occur at :
breakpoints of piecewise functions
.
( x > 2) →
§
, Infinite : when theres an
asymptote ( HA or VA )
can occur at :
limflx ) → IN and/or Iim →
± a
§ ,
c. =/ 0
g- →
§ ,
c -1-0 breakpoints
Piecewise function
of a
✗ → a- ✗ → at
E. or
E. →
§
^ ^
^ 1 ^
1 1 1
1
I 1 Kainuu
l l
l
- l
l
µ%s8 l
l
s
l s
s l s
1 i I 1
Limits and discontinuities
limflx) -_ too f- 1- 1) =3
ex :
✗ → -
2-
'
I ^ Iimflx )= I
1
limflx )= -
N
✗ → O
-4
l ✗ → -2T
/ wñ#i - ME
limflx )= ☐ NE
HO )= /
I -2 -
-
- - - -
-
- -
- - - -
✗ → -2
limflx )= -
l
l ✗ → 2-
f- (2) = DNE
-
I
limflx )= -
I
Iimflx )=o ✗ → 2T
I ✗ → -
l
-
l -
l
tim f- ( ✗ 1=3 limflx )= -
l
l ✗ → 2
-
-2 ✗ → -
it
| Iim f- 1×1=0
-
limflx )= DNE ✗ → -00
I ✗ → -
l
-4
Iimfcx ) 2
-
l
1 f- (2) =3 __
✗→ oo
Find the discontinuities and their
type
:
11=-2 : infinite
✗ = -1 :
jump
11=2 : removable
DIRECT SUBSTITUTION
if f- ( X ) is continuous at X=a
,
then limfcx)= f- (a) .
To evaluate the
✗ → a
limit , plug in a .
If it
gives you a number like 2 orbs ,
that's
your
answer .
If
you get
8- ,
facto rise 14 methods ) see next pages .
If
you get § ,
it's either
-
do ,
so or DNE .
Evaluate the limits :
tim .
✗2+4=8
→
22+4 Iim ✗ + 4- =3
✗ → 5
✗ → 2
Iim ✗ +3=6 lim X -
3=0
✗ → 3×2-4 5 ✗ →
3×2-9 ☐
Iim ✗ +3 =I
2
✗
X2 -4
→
0
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