3.1 b - MATH
1081
 Wednesday,
January
26
 


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Unformatted text preview: MATH
1081
 Wednesday,
January
26
 
 Chapter
3
–
Section
1(continued)
 
 LIMITS
 
 Homework
#2
(due
1/31):
 Section
3.1
#8,
34,
38,
40,
42,
46,

 48,
50
 
 Clicker
Check­in:

Choose
any
letter
to
check
in
now.
 Last
time,
we
examined
the
notion
of
a
limit
be
consulting
the
 graph
of
the
function
at
the
specified
point.
 
 Today,
we
will
examine
how
to
evaluate
a
limit
with
just
an
 algebraic
expression
representing
the
function.



 
 Of
course,
the
result
still
expresses
the
same
idea
as
when
 looking
at
the
graph.

We
just
compute
the
result
in
a
different
 manner.
 
 x 2 − 25 Consider
the
function
 f ( x ) = .

At
 x = 5 ,
this
function
 x−5 does
not
exist.
 
 However,
if
we
plug
in
values
of
 x 
near
 5 ,
we
see
the
following
 results.
 
 x
 4.95 
 5
 5.05 
 5.1
 5.2 
 4.8 
 4.9 
 f ( x )
 9.95 
 DNE
 10.05 
 10.1
 10.2 
 9.8 
 9.9 
 
 
 According
to
this
table
of
values,
what
are
the
values
of
 f ( x ) 
 tending
towards
for
 x 
closer
and
closer
to
 5 ?
 x 2 − 25 In
the
case
of
 f ( x ) = ,
when
 x = 5 
we
have
the
problem
 x−5 of
division
by
 0 .


 
 However,
for
any
 x ≠ 5 ,
we
have
the
common
(non‐zero)
factor
 of
( x − 5 ) 
that
can
be
cancelled
out
to
yield
( x + 5 ) .
 
 So,
although
 f ( 5 ) 
doesn’t
exist,
 
 x 2 − 25 ( x − 5 ) ( x + 5 ) = lim x + 5 = 10 .
 lim = lim ( ) x→ 5 x − 5 x→ 5 x→ 5 x−5 
 There
are
actually
rules
for
computing
limits
that
essentially
 boil
down
to
the
idea
that
MANY
functions
are
nice
and
the
 value
of
the
limit
at
 x = a 
is
actually
equal
to
the
value
of
the
 function
at
 x = a .
 
 So,
if
we
can
compute
 f ( a ) 
by
plugging
in
 a 
to
the
function
and
 get
a
real
number,
then
that
will
be
the
value
of
the
limit
of
 f ( x ) 
as
 x → a .
 
 However,
if
evaluating

 f ( a ) 
results
in
an
illegal
operation,
like
 division
by
0,
we
cannot
be
so
hasty
to
draw
a
conclusion
as
 x 2 − 25 the
example
with
 f ( x ) = 
showed.
 x−5 This
is
from
page
 155
in
your
text.

 Example:
 Compute
the
limit
or
determine
that
it
does
not

 exist.
 
 1. lim 3x 2 − 5 x + 1
 x→ 2 2. 3. 4. 5. x → −3 lim 2 x 
 x2 − 2x + 1 
 lim 2 x →1 x −1 ex ( x + 4 ) 
 lim x → − 4 x 2 + 8 x + 16 lim h→ 0 x+h − h 
 h What
issue(s)
do
we
run
into
when
computing
 lim x ?

 x→ 0 Or
 lim x ?
 x → −8 
 
 
 Do
we
have
the
same
issue(s)
when
computing
 lim 3 x ?
 x→ 0 Or
 lim 3 x ?
 x → −8 
 As
for
computing
limits
as
 x 
goes
to
 +∞ 
or
 −∞ ,
we
have
to
 consider
what
happens
to
the
expression
defining
the
function
 when
 x 
is
VERY
large
in
magnitude.
 
 For
polynomial
and
rational
functions,
we
can
use
the
idea
that
 the
term(s)
with
the
largest
exponent
will
be
so
much
larger
 than
all
other
terms
when
 x 
gets
big,
that
none
of
the
other
 terms
even
matter.
 
 The
book
describes
a
method
of
dividing
by
the
highest
power
 of
 x .

We
can
use
a
more
intuitive
method
just
by
inspecting
 and
isolating
the
terms
with
the
largest
exponents,
since
when
 the
value
of
 x 
gets
VERY
large,
the
term
with
the
largest
 exponent
will
dominate
all
the
otherterms.
 
 However,
if
you
prefer
a
specific
set
of
steps
to
follow,
 reference
the
“Finding
Limits
At
Infinity”
box
on
page
163.
 
 Example:
 Compute
the
limit
or
determine
that
it
does
not

 exist.
 
 1. lim 4 x 5 − 9 x 3 + 12 x + 1000 
 x→ ∞ 2. 3. 4. 5. x → −∞ lim − 2 x 6 + 25 x 5 + 7 x 4 − 90 x 3 + 1
 3x 2 − 5 x + 7 
 lim x → ∞ 9 x 3 + 17 x 5 x 5 − 4 x 2 + 9 x − 10 
 lim 4 3 2 x → −∞ 3x + 8 x − 2 x + x + 12 4 x 4 − 5 x 3 + 2 x 2 − 3x + 1 
 lim 4 x→ ∞ 2 x + 200 Clicker
Check­out:

Choose
any
letter
to
check
out
now.
 
 In
recitation
tomorrow
is
Quiz
1,
covering
Sections
1.1
&
2.1.
 
 Next
week
–
Sections
3.2
&
3.3.
 ...
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This note was uploaded on 05/01/2011 for the course MATH 1081 taught by Professor Johanson during the Spring '08 term at Colorado.

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