Download Trigonometric Limits

Trigonometric Limits
We will deal here with a set of problems posed at the end of Section 3.3. In order
to find the derivatives of sin x and cos x , using the definition of derivative, we
needed to prove the limits
sin x
= 1
x→0 x
lim
and
lim
x→0
cos x − 1
= 0 .
x
We are now asked to use these as the “tools” for calculating a set of limits involving
trigonometric functions on Problems 39 - 48.
€
39)
€
lim
x→0
sin 3x
x
There is really nothing special about the variable x in this limit expression. Just
as a letter can stand in for a number in algebra, a function u(x) can be used in place of
a variable x in an expression, provided that the expression is still meaningful. So
saying, we can extend our “limit law” above to read
sin u
= 1
u→0 u
lim
with u being some function of x . Pertinent to this problem, then, we can say that
sin 3x
sin 3x
= 1 , since x certainly goes
= 1 , from which it follows that lim
€
x → 0 3x
3x → 0 3x
lim
to zero if 3x does.
€
We now need to adapt this to the stated limit expression, which we can
accomplish by multiplying the numerator and denominator by 3 and applying the limit
€
laws appropriately:
lim
x→0

sin 3x
3 sin 3x
sin 3x
sin 3x 
= lim
⋅
= lim 3 ⋅
= 3 ⋅  lim
 = 3 ⋅1 = 3 .
x→0 3
x→0
 3x → 0 3x 
x
x
3x
We can generalize this example into a result for the limit
sin mx
= 1, m ≠ 0 , n ≠ 0
x →0
nx
lim
€
,
by using
sin mx
= 1 ⇒
mx → 0 mx
lim
€
sin mx
= 1
x → 0 mx
lim
and manipulating the limit expression
sin mx
sin mx
1 sin mx
m
= lim n
= lim n ⋅
=
x →0
x →0
nx € x → 0 ⋅ mx
mx
n
m
m
lim
€

sin mx 
m
m
⋅  lim
⋅1 =
 =
 x → 0 mx 
n
n
.
40)
€
lim
x→0
sin 4 x
sin 6x
Since there are no linear terms in this ratio at all, we will have to introduce them
ourselves. This will break up the limit expression into terms that we know how to
evaluate:
lim
x→0
=
€
 sin 4 x
 sin 4 x 4 x   6x
sin 4 x
1 
6x
4 x/ 
= lim 
⋅
⋅  = lim 
⋅
⋅


x→0  4x
x→0  4x
sin 6x
1   sin 6x 6x 
sin 6x 6 x/ 
sin 4 x
6x
sin 4 x
2
2
⋅ lim
⋅ lim
= ⋅ lim
⋅
3 4 x → 0 4 x 6x → 0 sin 6x
3 x→0 4x
We can make a generalization here for
€
lim
x→0
1
sin 6x
lim
x →0 6x
=
2
1
2
⋅1 ⋅ =
.
3
1
3
sin mx
, m≠0, n≠0 ,
sin nx
similar to the one we made from Problem 39:
lim
x→0
 sin mx
sin mx
nx
mx/ 
sin mx
nx
m
= lim 
⋅€
⋅
⋅ lim
 = n ⋅ lim
x → 0  mx
mx → 0 mx
nx → 0 sin nx
sin nx
sin nx nx/ 
€
x →0
41)
€
sin mx
1
m
m
1
m
= n ⋅ lim
⋅
= n ⋅1 ⋅ = n
sin
nx
1
x → 0 mx
lim
nx
tan 6t
t → 0 sin 2t
€
lim
We will break up this limit expression in a manner similar to what we used in the
previous Problem, keeping in mind the definition of the tangent function:
lim
t →0
tan 6t
sin 6t cos 6t
sin 6t
1
= lim
= lim
⋅
t →0
t → 0 sin 2t cos 6t
sin 2t
sin 2t
.
From here, we proceed in the way we have before:
lim
€
t →0
 sin 6t
sin 6t
1
2t
6t 
1
⋅
= lim 
⋅
⋅  ⋅ lim
t → 0  6t
sin 2t cos 6t
sin 2t 2t  6t → 0 cos 6t
€
€
sin 6t
2t
1
⋅ lim
⋅ lim
6t → 0 6t
2t → 0 sin 2t 6t → 0 cos 6t
= 3 ⋅ lim
€
.
sin 6t
2t
1
1
⋅ lim
⋅
= 3 ⋅1 ⋅1 ⋅ = 3 .
1
t → 0 6t
t → 0 sin 2t lim cos 6t
t →0
= 3 ⋅ lim
lim
42)
θ →0
cos θ − 1
sin θ This is also a limit expression that needs to be re-written in terms of ratios for
which we do know limits (the two limits shown at the beginning of these notes):
€
lim
θ →0
cos θ − 1
cos θ − 1 θ
cos θ − 1
θ
= lim
⋅
= lim
⋅ lim
θ →0
θ →0
θ → 0 sin θ
sin θ
θ
sin θ
θ
= lim
€
lim
43)
€
€
θ →0
θ →0
cos θ − 1
1
1
⋅
= 0⋅ = 0 .
sin
θ
θ
1
lim
θ →0 θ
sin( cos θ )
sec θ
This limit requires a somewhat more sophisticated approach, since the
expression involves a composite function. We will need to make a substitution for the
cosine function, u = cos θ , which then requires us to make two transformations of the
limit expression. One is to re-write the denominator as
sec θ =
1
1
= u
cos θ
The other change is to replace the limit variable
we have u = cos
lim
θ →0
.
θ with u : with θ approaching zero,
θ approaching cos 0 = 1 . Our limit can then be evaluated as
€ θ)
sin( cos
sin( u )
= lim
= lim u sin( u ) = 1 ⋅ sin 1 = sin 1 .
u →1 1 u
u →1
sec θ
This is the sine of the angle 1 radian, so it does not have a simpler form.
€
44)
sin 2 3t
lim
t →0
t2
This is another limit expression we must evaluate by factoring it first:
€
lim
t →0
sin 2 3t
sin 3t sin 3t
sin 3t
sin 3t
=
lim
⋅
=
lim
⋅
lim
= 3⋅ 3 = 9 .
t →0
t →0
t →0
t
t
t
t
t2
This limit expression could also be written using the “limit laws” as
€
lim
t →0
€
2
 sin 3t 2 
sin 2 3t
sin 3t 
=
lim
=
lim




t →0  t

t → 0 t 
t2
.
This remaining limit can be evaluated using our result from Problem 39, or the
generalization we described there.
lim
45)
€
θ →0
sin θ
θ + tan θ
This limit is not convenient to evaluate in its given form; instead, we will work
with the reciprocal of the expression and use the “limit law”
 1 
1
lim 
 = lim u(x)
x →a  u(x) 
x →a
.
So saying, we have
lim
θ →0
sin θ
1
1
1
= €
=
=





θ + tan θ
tan θ
sin θ cos θ 
θ + tan θ
lim 
lim  θ +
lim  θ +



sin θ 
sin θ 
θ → 0  sin θ 
θ → 0  sin θ
θ → 0  sin θ
=
€
1


lim  θ + 1 
cos θ 
θ → 0  sin θ
=
1
lim θ + lim 1
sin θ
θ → 0 cos θ
θ →0
=
1
1
sin θ
lim
θ →0 θ
+
1
lim cos θ
θ →0
=
€
€
46)
€
lim
x→0
1
1 1
+
1 1
=
1
1
=
.
1+1
2
sin (x 2 )
x
As with Problem 43, because the limit expression uses a composite function, we
need to make the substitution u = x2 . The denominator then becomes x → √u and the
limit variable is transformed into √u → 0 and so into u → 0 . This leads us to
sin (x 2 )
sin (u)
lim
= lim
.
x→0
u→0
x
u
We still aren’t quite ready to work with this, since it involves an expression for
which we don’t know the limit. If we now make a suitable multiplication of the
numerator and
€ denominator, we obtain
lim
u→0
€
sin (u)
= lim
u→0
u

 
u sin (u)
sin (u)  
sin (u) 
⋅
= lim  u ⋅
 =  lim u  ⋅  lim
 = 0 ⋅1 = 0 .
u→0 
 u → 0 u 
u  u → 0
u
u
lim
47)
x →π /4
€
1 − tan x
sin x − cos x We find that, by a straightforward application of the “limit laws” that this
expression gives us a ratio of
π
1 − tan 4
1 − 1
0
= " " ,
π
π =
2
2
0
sin 4 − cos 4
− 2
2
so we must find some way to manipulate this limit expression to see if we can obtain
something other than an indeterminate ratio. If we multiply the numerator and
denominator by cos x , we find that we can then factor the resulting ratio:
€
lim
x →π /4
 sin x 
1 − 

 cos x  cos x
⋅
=
sin x − cos x cos x
lim
x →π /4
=
€
x →π /4
48)
€
€
lim
− (sin x − cos x)
=
(sin x − cos x) ⋅ cos x
lim
x →1
lim
x →π /4
cos x − sin x
(sin x − cos x) ⋅ cos x −1
−1
−1
2
=
=
= −
= − 2
π
 2
cos x
2
cos 4
 2 
 
sin (x − 1)
x2 + x − 2
This is another limit expression which uses a composite function; in addition, we
have a more complicated denominator, but one which can be factored. If we perform
the factorization first, this becomes
lim
x →1

sin (x − 1)
sin (x − 1)
sin (x − 1)  
1 
= lim
=  lim
⋅  lim
 .

2
x →1 (x − 1) (x + 2)
 x →1
x − 1   x →1 x + 2  x +x−2
We now deal with the composite function by using the substitution u = x – 1 , which will
also change the limit variable to have u approach zero:
€
lim
x →1
€

sin (x − 1)
sin u  
1 
1
1
=
lim
⋅
lim

 = 1⋅ 1+ 2 = 3 .


2
 u → 0 u   x →1 x + 2 
x +x−2
Later in this course (Section 4.4), we will learn about a more powerful method
which will (usually) allow us to work out this sort of limit problem more quickly.
Nonetheless, it is useful to know about different ways of manipulating limit expressions
in order to make them easier to evaluate.
G. Ruffa
original handout – October 2005
revised 20-25 August 2009
.