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2 – 5B: Special Trig Limits
!
Math 400
This section focuses entirely on the limits of trigonometric functions. The use of Theorem 1, Theorem
2 and the indeterminate cases of Theorem 1 are all considered.
The limit of the function f (x) as x approaches a
from the left OR right side where there is a point at ( a , f (a))
and f (x) is continuous in the “neighborhood of x = a
Theorem One:
lim
f (x) = f (a) if f (a)∈REALS AND f (x) is continuous in the “neighborhood of x = a
x→ a
In many cases the limit of f (x) as x approaches a can be found by the use of Theorem 1.
Example 1!
Example 2!
lim
sin(x)
x→π 2
2x
sin(x)
is continuous in
2x
the neighborhood of π 2 so
lim
x →π 2
=
sin(x) sin(π 2)
=
2x
2(π 2)
lim
cos(3x)
x→π 3
6x
cos(3x)
is continuous in
6x
the neighborhood of π 3 so
!
1
π
lim
cos(3x) cos( 3(π 3))
=
x→π 3
x
6(π 3)
=
cos(π ) 0
=
=0
2π
2π
Example 3
lim 1− cos(x)
x→π
x
1− cos(x)
is continuous in
x
the neighborhood of π so
lim
x →π
1− cos(x) 1− cos(π ) 2
=
=
x
π
π
Lecture 2 – 5B !
Page 1 of 12!
© 2016 Eitel
Limits involving Vertical Asymptotes of the basic trigonometric functions
In some cases the graph of the basic trigonometric function has a vertical asymptote at x = a. The
cases where the basic trigonometric function has a vertical asymptote at x = a requires that you know
something about the graph of trigonometric function in the neighborhood of x = a
Several of the infinite periods for y = tan ( x )
y
−2π
−5π
2
−π
−3π
2
π
−π
2
π
2
Vertical Asymptotes occur at x = ...,
x
2π
3π
2
5π
2
−7π −5π −3π −π π 3π 7π
,
,
, ,
,
,....
2
2
2 2 2 2 2
Each of the Vertical Asymptotes is an Odd Asymptote
Inspecting the graph allows us to answer the following limits:
1.
lim
3π − tan (x) = +∞!
x→
2
lim
3π + tan (x) = −∞!
2.
x→
2
lim
3. x → 3π tan (x) = DNE
2
Several of the infinite periods for y = cot ( x )
y
−π
2
−3π
2
−2π
−π
π
2
3π
2
π
x
5π
2
2π
3π
Vertical Asymptotes occur at x = ..., − 4π , − 2π , 0, 2π , 4π , ...
Each of the Vertical Asymptotes is an Odd Asymptote
Inspecting the graph allows us to answer the following limits:
4.
lim
cot (x) = +∞!
x → −π+
Lecture 2 – 5B !
5.
lim
cot (x) = −∞!
x → 3π −
Page 2 of 12!
6.
lim
cot (x) = DNE
x→ 0
© 2016 Eitel
Several of the infinite periods for y = sec ( x )
y
−2π
2π
1
–1
−π
−5π
2
x
−3π
2
π
−π
2
π
2
Vertical Asymptotes occur at x =
...,
3π
2
5π
2
−7π −5π −3π −π π 3π 7π
,
,
, ,
,
,....
2
2
2 2 2 2 2
Each of the Vertical Asymptotes is an Odd Asymptote
Inspecting the graph allows us to answer the following limits:
7.
lim
π − sec (x) = +∞!
x→
2
8.
lim
π + sec (x) = −∞!
x→
2
lim
9. x → π sec (x) = DNE
2
Several of the infinite periods for y = csc ( x )
y
−3π
2
5π
2
π
2
1
x
−π
2
−2π
–1
3π
2
−π
π
2π
3π
...,−3
π
,
−
3
π
,
−
1
π
,
0
,1
π
,2
π
,3π ,4π ,....
Vertical Asymptotes occur at x =
Inspecting the graph allows us to answer the following limits:
lim
− csc (x) = −∞
10. x → 2π
!
Lecture 2 – 5B !
lim
+ csc (x) = +∞
11. x → 2π
!
Page 3 of 12!
lim
csc (x) = DNE
12. x → 2π
© 2016 Eitel
Limits involving Vertical Asymptotes of the basic trigonometric functions
In some cases the graph of the basic trigonometric function has a vertical asymptote at x = a. You
can use Theorem 2 to determine the limit of the function as x approaches and there is a vertical
asymptote at x = a.
Theorem 2A
The limit of the function as x approaches a
from the left OR right side of a vertical asymptote at x= a
lim
f (x)
= +∞ or − ∞
−
x→a
g(x)
based on the sign of f (a − Δx)
and
lim
f (x)
= +∞ or − ∞
+
x→a
g(x)
based on the sign of f (a + −Δx)
The limit of the function as x approaches a from the left or right side of a vertical asymptote at x = a
is either +∞ or − ∞ . To determine the sign of ∞ , plug in a number “close to a“ on the side
indicated and find the sign of that number.
Example 1
lim
tan (x)
x → (π / 2) −
Example 2
lim
tan (x)
x → (π / 2) +
tan (x) has a vertical asymptote at π / 2
tan (x) has a vertival asymptote at π / 2
determine the sign of tan (x − Δx)
determine the sign of tan (x + Δx)
so the sign is positive
so the sign is negitive
lim
tan (x) = +∞
x → (π / 2) −
lim
tan (x) = −∞
x → (π / 2) +
tan((π / 2) − Δx ) is in the first quadrant
tan((π / 2) + Δx ) is in the second quadrant
!
Example 3
lim
cot (x)
x → π−
Example 4
lim
cot (x)
x → π+
cot (x) has a vertical asymptote at π
cot (x) has a vertical asymptote at π
determine the sign of cot (x − Δx)
determine the sign of cot (x + Δx)
cot (π − Δx ) is in the second quadrant
so the sign is negitive
cot (π + Δx ) is in the third quadrant
so the sign is positive
lim
cot (x) = −∞
x → π−
lim
cot (x) = +∞
x → π+
Lecture 2 – 5B !
!
Page 4 of 12!
© 2016 Eitel
The Indeterminate Cases for trigonometric functions
lim
f (x)
f (a)
produces
= a Real Number This
x → a g(x)
g(a)
f (a)
result allows us to state that the limit as x approaches a is
y. Sometimes the use of Theorem 1
g(a)
lim
f (x)
f (a)
zero
to find the limit
produces a value of
=
. We call this the Indeterminate
x → a g(x)
g(a)
zero
Many times the use of Theorem 1 to find the limit
case for Theorem 1.
There are 2 outcomes for the indeterminate case.
Either the the limit as x approaches a is a real number value or the graph of the function at x = a has
a vertical asymptote at x = a and the limit of the function as x approaches a from the left or right side
of a vertical asymptote at x = a is either +∞ or − ∞.
zero
cases describes a “holeʼ or a vertical asymptote in the graph?
zero
lim
f (x)
f (a) zero
If the use of Theorem1 for the
produces
=
x → a g(x)
g(a) zero
then
f (x)
h(x)
"reduce"
to
and RETEST x = a,
g(x)
k(x)
one of two things MUST happen
How do you determine if the
Case 1(hole)
If
lim
x → a+
h(a)
is a REAL number then there is a "hole" in the graph at x = a and y =
k(a)
and
lim
f (x)
= the limit of the reduced function
x → a g(x)
lim
lim
h(x) h(a)
h(x) h(a)
=
and
=
and
−
x → a k(x) k(a)
x→a
k(x) k(a)
h(a)
k(a)
h(x) h(a)
=
k(x) k(a)
OR Case 2 (vertical asymptote)
h(a)
non zero number
If
is a
then there is a "vertical asymptote" in the graph at x = a
k(a)
zero
and
lim
lim
f (x)
f (x)
= +∞ or − ∞ and
= +∞ or − ∞
−
+
x→a
x→a
g(x)
g(x)
lim
f (x)
= +∞ or − ∞ or DNE
x → a g(x)
Determine each limit if it exists.
Lecture 2 – 5B !
Page 5 of 12!
© 2016 Eitel
Theorem
lim sin x
lim
x
= 1 and
=1
x→ 0
x → 0 sin x
x
sin x
x
or
and that
x
sin x
the limit be taken as x → 0. The limit of other functIons cannot be taken using this theorem.
This limit theorem requires that you have an exact match to the functions
Example 1!
Example 2!
lim
5x
x → 0 sin(x)
lim sin(x)
x→ 0
2x
=
=
lim 1 sin(x)
•
x→ 0 2
x
1 lim sin(x)
•
2 x→ 0
x
1
1
= •1 =
2
2 !
=
lim 5
x
•
x → 0 1 sin(x)
=
5 lim
x
•
1 x → 0 sin(x)
=
5
5
•1 =
1
1
Theorem
If ax → 0 then x → 0
where a is a nonzero real number
Theorem
lim sin(ax)
lim
ax
= 1 and
= 1 where a is a nonzero real number
x→ 0
x → 0 sin ax
ax
This limit theorem requires that you have an exact match to the functions
sin( ax))
ax
or
and
ax
sin( ax )
that the limit be taken as x → 0 . The limit of other functIons cannot be taken using this theorem.
Example 3!
Example 4!
lim sin(8x)
x→ 0
8x
lim
3x
x → 0 sin(3x)
lim sin(ax)
= 1 a ∈nonzero real
x→ 0
ax
so
lim sin(8x)
=1
x→ 0
8x
!
lim sin(ax)
= 1 a ∈nonzero real
x→ 0
ax
so
lim
3x
=1
x → 0 sin(3x)
Lecture 2 – 5B !
Page 6 of 12!
© 2016 Eitel
Theorem
lim sin(ax)
lim c
ax
=
•
= 1 where cand a are nonzero real numbers
x→ 0
x → 0 c sin(ax)
ax
In English: You can multiply
sin(ax)
ax
c
or
by or a from of 1 and not change the limit.
ax
sin( ax )
c
Theorem
lim
lim
a • f (x) = a •
f (x) where a is a nonzero real number
x→ 0
x→ 0
In English: You can factor out a constant from the limit and not change the limit.
Example 5!
Example 6
lim
x
x →0 sin(7x)
lim sin(4 x)
x→ 0
x
multiply
sin(4 x)
4
by
x
4
multiply
lim
4 • sin(4 x)
x→ 0
4x
x
7
by
sin(7x)
7
lim
7x
x →0 7 sin(7x)
!
lim
sin(4 x)
4•
x→ 0
4x
=
!
lim 1
7x
•
x →0 7 sin(7x)
lim sin(4 x)
4
•
1 x → 0 4x
= 4 •1 = 4
Lecture 2 – 5B !
1 lim
7x
•
7 x →0 sin(7x)
=
1
1
•1 =
7
7
Page 7 of 12!
© 2016 Eitel
Example 7!
Example 8
lim 2sin(4 x)
x→ 0
3x
lim
x3
x → 0 sin 3 (2x)
rewrite as
lim 2 sin(4 x)
•
x→ 0 3
x
2 sin(4 x)
4
multiply •
by
3
x
4
rewrite as
lim
x
x
x
•
•
x → 0 sin(2x) sin(2x) sin(2x)
lim 8 sin(4 x)
•
x→ 0 3
4x
8 lim sin(4 x)
•
3 x→ 0
4x
=
8
8
•1 =
3
3
Lecture 2 – 5B !
multiply by
!
2• 2• 2
8
!
lim 1
2x
2x
2x
•
•
•
x → 0 8 sin(2x) sin(2x) sin(2x)
1 lim
2x
2x
2x
•
•
•
8 x → 0 sin(2x) sin(2x) sin(2x)
=
1
1
•1•1•1 =
8
8
Page 8 of 12!
© 2016 Eitel
Theorem
lim sin(ax)
lim x
ax
=
•
=1
x→ 0
x → 0 x sin(ax)
ax
x
x
In English: You can multiply by and not change the limit because
= 1 for all x ≠ 0
x
x
lim sin(ax)
x
The
requires x to approach BUT NOT REACH 0 so = 1 as x → 0
x→ 0
ax
x
Example 9!
Example 10
lim
2
x → 0 sin(2x)
lim sin(4 x)
x→ 0
4
multiply
sin(4 x)
x
by
4
x
multiply
lim
x • sin(4 x)
x→ 0
4x
=
lim
2x
x → 0 x • sin(2x)
lim
lim sin(4 x)
x •
x→ 0
x→ 0
4x
= 0 •1 = 0
Lecture 2 – 5B !
2
x
by
sin(2x)
x
=
!
lim 1
lim
2x
•
x → 0 x x → 0 sin(2x)
= DNE •1 = DNE
Page 9 of 12!
!
© 2016 Eitel
Example 11!
Example 12
lim
sin(4 x)
x → 0 2sin(5x)
lim 1− cos(4 x)
x→ 0
sin(3x)
rewrite as
lim 1 sin(4 x)
1
•
•
x→ 0 2
1
sin(5x)
rewrite as
lim 1− cos(4 x)
1
•
x→ 0
1
sin(3x)
the sin(4 x) term needs a 4x in it's denominator
and the sin(5x) term needs a 5x in it's numarator
sin(4 x)
4x
1
5x
multiply
by
and
by
1
4x
sin(5x)
5x
the 1− cos(4 x) term needs a 4x in it's denominator
and the sin(3x) term needs a 3x in it's numarator
1− cos(4 x)
4x
1
3x
multiply
by
and
by
1
4x
sin(3x)
3x
=
lim 1 4 x • sin(4 x)
5x
•
•
x→ 0 2
4x
5x • sin(5x)
=
lim
4 x • (1− cos(4 x))
3x
•
x→ 0
4x
3x • sin(3x)
=
lim 1 4 x sin(4 x)
lim sin(4 x)
•
•
•
x → 0 2 5x
x→ 0
4x
4x
=
lim
lim
4x 1− cos(4 x)
3x
•
•
x → 0 3x
x → 0 sin(3x)
4x
=
lim 2 sin(4 x) lim sin(4 x)
•
•
x→ 0 5
x→ 0
4x
4x
=
lim
lim
4 1− cos(4 x)
3x
•
•
x→ 0 3
x → 0 sin(3x)
4x
=
2
2
•1•1 =
5
5
=
4
• 0 •1 = 0
3
!
!
Lecture 2 – 5B !
Page 10 of 12!
© 2016 Eitel
Example 13!
Example 14
lim sin 2 (4 x)
x→ 0
5x 2
lim 1− cos(4 x)
x→ 0
sin(3x)
lim 1 sin(4 x) sin(4 x)
•
•
x→ 0 5
x
x
rewrite as
lim 1− cos(4 x)
1
•
x→ 0
1
sin(3x)
each of the x's in the denominator needs to be a 4x
multiply by
=
16
16
=
16 4 • 4
the 1− cos(4 x) needs a 4x in it's denominator
and the term needs a 3x in it's numarator
3x
multiply by
4x
lim 1 4 sin(4 x) 4 sin(4 x)
•
•
x→ 0 5
4x
4x
lim 16 sin(4 x) sin(4 x)
=
•
•
x→ 0 5
4x
4x
=
16 lim sin(4 x) lim sin(4 x)
•
•
x→ 0
5 x→ 0
4x
4x
=
16
16
•1•1 =
5
5
=
lim 1− cos(4 x)
3x
•
x→ 0
4x
sin(3x)
=
lim 1− cos(4 x)
lim
3x
•
x→ 0
x → 0 sin(3x)
4x
= 0 •1 = 0
!
!
Lecture 2 – 5B !
Page 11 of 12!
© 2016 Eitel
Limits of the trigonometric functions as x → ± ∞
All 6 trigonometric functions are periodic functions. This means that the basic one period for each
function repeats over and over again as the x values approach infinity. For this reason the limit for
each function as the x values approach positive or negative infinity x → ± ∞ DO Not Exist.
Inspecting the graphs of the trigonometric functions allows us to answer the following limits:
13.
lim
sin (x) = DNE !
x → ±∞
14.
lim
cos (x) = DNE !
x → ±∞
15.
lim
tan (x) = DNE
x → ±∞
16.
lim
csc (x) = DNE !
x → ±∞
17.
lim
sec (x) = DNE !
x → ±∞
18.
lim
cot (x) = DNE
x → ±∞
Theorem
lim sin x
=1
x→ 0
x
This limit theorem requires that you have an exact match to the functions
taken as x → 0 . No other functions limit can be taken using the theorem.
You cannot expect this theorem to work for
lim sin( f (x))
x→0
f (x)
sin x
and that the limit be
x
where f(x) is a function other than y =x
The theorem listed above cannot be used to answer limits like the ones listed below.
( )
2
lim sin x
x→ 0
x2
Lecture 2 – 5B !
( )
!
x
lim sin e
x→ 0
ex
!
Page 12 of 12!
lim sin(ln(x))
x→ 0
ln(x)
!
© 2016 Eitel