Class notes 201-NYA-05

Limits
- value function approaches as the input approaches some value
that a
lim
1 a function f has a limit Las X approaches a , written xeaf(x) = L if the

close to the number
Las we please by taking x
sufficiently close to a lon eit

limit definition : lim f(x+h) f(x) -




neo
h



ex)
im 4x2 + lim2x +im 2
X-2 X-2


=4 [limx] + 2 limx+ lim 2
X+ 2
X92Xe2

=
4[232 + 2(2) + 2

=
16 + 4 + 2 =
22 : The limit exists & converges to 22




limits
Evaluating ~ ratio of 2 p

· substitution property : if f(x) is a polynomial or a rational function

< then lim f(x) F(a) lother functions benefit this property)
may
: =

Xe A


< lim DN
& of limits 1)
2
types : xea
limits can
equal a number ,
infinity or
m
2)x > 10




/im ==DNE ex3) lim
ex1) ex 2) lim -
X = 0 = DNE
10gaX
X- 0


"
& DNE
" -



DNE ⑧
-




&
!
E

b
[ 2
b
[



1 both sides aren't equal


Evaluating limit techniques
C First case : lim P(x) & the limit will evaluate to 0/0 when .
X=a Then ,
x+
aQ(x)


1) if P & Q are
polynomials & P(a)=Q(a) = 0 , then X-a is a factor of bot

<lim P(x) (a)(g(x)) g(x) division if ca
xaQ(x)
= =
(might have to use long
(a)(k(x))k(X)

2) if P & Q are not polynomials & one or both has a radical , we rationaliz


3) if P & Q are not polynomials & contains a fraction on either the

, =lim x2 3x -1 (x + 2) x2 + 2x (x -1)(x + 3) (x + 3)) =
+ 3
( 4
-
-


= = =
-
-




Xe1- -
(x -

1) -
(X -

1) -
(x1) : the lim D
X-

= lim x2 + 3x -1 -
(x + 2) =
x2 + 2x -
3 =
(x)(x + 3) = (x + 3) = 4
X- It (X 1) - X -
1 -1


ex 3) lim X -

2x-1 .
x + 2x -
1 .
x+3 + 2

x+1x + 3
-


2 X+ 2x -
1 X+ 3 +2



= lim (x= 2x + 1) .
(x + 3 + 2)
X-1
X+ 3 -

4 X + 2x -
1


= lim (x-1)((x) .
X+ 3 + 2
X- /
(1) X+ 2X -
1



= lim(x -1)(x + 3 + 2) = 0
X91 X+ 2X - 1




Limits


here is a problem using long division :




ex : lim x3 + 6x2 + 4x-1
Xe -
It x2 + 2x + 1
x2 + 5x -
1
lim x3 + 6x2 + 4x 1
X + 1x3 + 6x2 +
=
-




> 4X 1
It
-




Xe
(X + 1)2
-




- x3 + x2 d
=lim(xF1)(y2 + 5x 1) -

05x2 + 4x
X- -
It
+ 1)2
(x- 5x2 + 5xx
-




0 -
X -
1
lim
5
= =
-
0(V A
--X 1
-

= .
-



Xe -
It
O



Vertical Asymptotes
& the line x=a is called a V A .
.
of the curve f(x) if at least one of the
followi
f(x) f(x) = f(x)
1)
lim
= 3)
lim 5) lim d
=



Xe at



2) lim 4) f(x) = f(x)
f(x) = d
lim 6) lim -
-
=


X- a xe at



:
only one of the above conditions must be met for there to be a V A
. .
at
( sometimes ,
V A . . occur when the limit evaluates to
& we've already seen functions with V A .
:




im logaX=-D
, a clim1
Xe0X2
= @ (both sides go to
infinity

, =

Xim 3
+ ((x 3Y(x - -
1)
-




zix 3) -




x(x 1) x(x) (always have
im [ 3 simplify
2x - -




=
-


- = -
to to see
+
(x 3)2(x 1)
- -

2(*3)(x 1) -




6x2 + X -
1
1) lim X+ 2 2) lim
t
4x2 4X 3
X- 8 X+
1
- -



tanx
-




L= lim 10) + 2 =
2
this means we have =lim (3x -

1)(x
+ 1)
&
X 90 O to check one-sided limits Xe -



1 (2x -


3)()
↓
= lim 3x -
1 =
3) 2) -
1
=
+
lim X+ 2 xe
= - D - 2x -
3 2( -

2) -

3
xeo-tanx >the graph of tanx tells
lim
us that tanx = 0-
-



X 90


X + 2 +
R = lim < lim tanx = o = d
Yeot tanx X - Ot




.
im Xt2 does not exist




Limits : Horizontal Asymptotes (H .
A) .




~ H A . . occur when the limit as X approaches infinity or
negative infinity g
The line y=L is called a H A
. . of the curve y=f(x) if :




lim f(x) L

=
, or lim f(x) =
L
X- * X- -
D




A
graphically
< it would look like A




-
,
:



---------------

S >
u
S 7
re
v


a

when X a proaches W W
+
& , there are no

one-sided limits lim f(x) = L lim f(x) = 1
X- -

D Xe



<
these limits will evaluate to 8 or - (indeterminant)


① When we obtain , we would divide all values in numerator & denominato

give us a value (c)
= if the highest power is in the numerator , the limit is I &
< if the
highest power is in the denominator , the limit is O

② if we have
o-o , we usually :
Mark: We will use a lot of