calc 1.6(10)
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7/23/2019 calc 1.6(10)
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Continuity of Trig and InverseFunctions
Objective: To use limits to define
continuity in trig/inverse functions
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7/23/2019 calc 1.6(10)
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Theorem 1.6.1
If c is any number in the natural domain ofthe stated trigonometric function, then
cxcx
sinsinlim =
cxcx
tantanlim =
cxcx
coscoslim =
cxcx
cotcotlim =
cxcx
secseclim =
cxcx
csccsclim =
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7/23/2019 calc 1.6(10)
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!am"le 1
Find the limit
1
1coslim2
1 x
xx
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7/23/2019 calc 1.6(10)
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!am"le 1
Find the limit
The limit of the cosine is the cosine of thelimit.
1
1coslim2
1 x
xx
( ) 2cos)1(limcos1
1limcos
1
1coslim
1
2
1
2
1
=+=
=
xx
x
x
x
xxx
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7/23/2019 calc 1.6(10)
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Theorem 1.6.#
If f is a one-to-one function that iscontinuous at each point of its domain,then f -1 is continuous at each point of its
domain, that is f -1is continuous at eachpoint of the range of f.
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7/23/2019 calc 1.6(10)
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!am"le $
Where is the function
continuous? 4
lntan)(
2
1
+=
x
xxxf
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7/23/2019 calc 1.6(10)
7/20
!am"le $
Where is the function
continuous?
A fraction is continuous where thenumerator and denominator arecontinuous and the denominator is not
ero.
4
lntan)(
2
1
+=
x
xxxf
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7/23/2019 calc 1.6(10)
8/20
!am"le $
Where is the function
continuous?
A fraction is continuous where thenumerator and denominator are continuousand the denominator is not ero.
The numerator is continuous for ! " #$why?% and the denominator is continuouse&erywhere
4
lntan)(
2
1
+=
x
xxxf
2x
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7/23/2019 calc 1.6(10)
9/20
!am"le $
Where is the function
continuous?
A fraction is continuous where the numeratorand denominator are continuous and thedenominator is not ero.
The numerator is continuous for ! " # $why?%and the denominator is continuouse&erywhere
The function f is continuous for ! " #, not '.
4
lntan)(
2
1
+=
x
xxxf
2x
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7/23/2019 calc 1.6(10)
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The %&uee'ing Theorem
(et f, g, and h be functions satisfying
for all ! in some open inter&al containing the
number c, with the possible e!ception thatthe ine)ualities need not hold at c. If g and hha&e the same limit as ! approaches c, say
then f also has this limit
as ! approaches c, that is
)()()( xhxfxg
Lxhxg cxcx == )(lim)(lim
Lxfcx
=
)(lim
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7/23/2019 calc 1.6(10)
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Theorem 1.6.(
1sin
lim)(0
= x
xa
x0
cos1lim)(
0=
x
xb
x
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7/23/2019 calc 1.6(10)
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!am"le )
Find*x
xx
tanlim0
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7/23/2019 calc 1.6(10)
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!am"le )
Find*x
xx
tanlim0
111cos
1lim
sinlim
cos
1sinlim
000==
=
xx
x
xx
x
xxx
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7/23/2019 calc 1.6(10)
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!am"le )
Find*x
xx
2sinlim
0
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7/23/2019 calc 1.6(10)
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!am"le )
Find*x
xx
2sinlim
0
2122
2sinlim2
2
2sin2lim
00===
x
x
x
x
xx
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7/23/2019 calc 1.6(10)
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!am"le )
Find*x
x
x 5sin
3sinlim
0
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7/23/2019 calc 1.6(10)
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!am"le )
Find*x
x
x 5sin
3sinlim
0
x
xx
x
x 5sin
3sin
lim0
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7/23/2019 calc 1.6(10)
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!am"le )
Find*x
xx 5sin
3sinlim
0
x
xx
x
x
xx
x
xx
5
5sin5
3
3sin3
lim5sin
3sin
lim00
=
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7/23/2019 calc 1.6(10)
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!am"le )
Find*x
xx 5sin
3sinlim
0
x
xx
x
x
xx
x
x
xx
x
xxx
5
5sin3
3sin
lim5
3
5
5sin5
3
3sin3
lim5sin
3sin
lim000
==
5
3
1
1
5
3=
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7/23/2019 calc 1.6(10)
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*ome+or,
+ection 1.
ages1'-1'
1-/ odd
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