M1271-Trig Limits - Lecture notes 4.7 PDF

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Stewart, Calculus: Early Transcendentals, 6th Edition...

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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

lim x→0

sin x = 1 x

lim

and

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

lim u→0

sin u = 1!! u

with u being some function of x . Pertinent to this problem, then, we can say that

lim 3x → 0

sin 3x sin 3x = 1 , since x certainly goes = 1 !!, from which it follows that lim x → 0 3x 3x

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 ⋅ = 3 ⋅  lim = lim 3 ⋅ = lim  = 3 ⋅1 = 3 . x→0 3 x→0 3x 3x → 0 3x  x x ! 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 sin mx  m m  m = lim n = lim n ⋅ = ⋅1 = ⋅  lim  = x →0 x →0 x →0 nx mx n n n x → 0 mx  ⋅ mx m lim

m

.

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 6x 4 x/  1  sin 4 x = lim  ⋅  ⋅ ⋅ ⋅  = lim   x→0  4x x→0 4x sin 6x sin 6x 6x/  1   sin 6x 6x 

sin 4 x 6x sin 4 x 2 2 ⋅ lim ⋅ lim ⋅ = ⋅ lim 3 x→0 4x 3 4 x → 0 4 x 6x → 0 sin 6 x

We can make a generalization here for

1 sin 6x lim x →0 6x

=

2 1 2 . ⋅1 ⋅ = 3 3 1

sin mx , m≠0, n ≠0 , sin nx

lim x →0

similar to the one we made from Problem 39:

lim x→0

 sin mx nx sin mx sin mx nx m/x  m = lim  ⋅ ⋅ ⋅ lim  = n ⋅ lim ! x → 0  mx mx → 0 mx nx → 0 sin nx sin nx sin nx n/x 

! !

41)

!

!

lim t →0

sin mx m ⋅ = n ⋅ lim x→0 mx

1 sin nx lim x →0 nx

m m 1 = n ⋅1 ⋅ = n . 1

tan 6t sin 2t

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 1 2t 6t  sin 6t 1 ⋅ ⋅ ⋅  ⋅ lim = lim  ! t → 0  6t sin 2t cos 6t sin 2t 2t  6t → 0 cos 6t

! !

= 3 ⋅ lim 6t → 0

sin 6t 2t 1 ⋅ lim ⋅ lim ! 2t → 0 6t → 0 cos 6t 6t sin 2t

! !

sin 6t 2t 1 1 = 3 ⋅1 ⋅1 ⋅ = 3 . ⋅ lim ⋅ cos 6t lim 1 t → 0 6t t → 0 sin 2t

= 3 ⋅ lim

t →0

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

!

θ cos θ − 1 cos θ − 1 θ cos θ − 1 = lim ⋅ = lim ⋅ lim θ →0 θ →0 θ → 0 sin θ θ θ sin θ sin θ ! = lim

!

43)

θ →0

!

lim θ →0

1 cos θ − 1 1 = 0⋅ = 0 . ⋅ θ 1 lim sin θ 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

θ with u : with θ approaching zero, θ approaching cos 0 = 1 . Our limit can then be evaluated as

The other change is to replace the limit variable we have u = cos

lim

θ →0

sin( cos θ ) sin( u ) = lim u sin( u ) = 1 ⋅ sin 1 = sin 1 . = lim 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)

lim t →0

sin 2 3t t2

This is another limit expression we must evaluate by factoring it first:

lim t→0

sin 3t sin 3t sin 3t sin 3t sin 2 3t = lim = lim ⋅ lim = 3⋅3 = 9 . ⋅ 2 t→0 t→0 t→0 t t t t t

This limit expression could also be written using the “limit laws” as

lim t→0

sin 2 3t t2

2  sin 3t  2  sin 3t  = lim   =  lim  t→0  t  t → 0 t 

.

This remaining limit can be evaluated using our result from Problem 39, or the generalization we described there. !

45)

lim 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 = = =      tan sinθ cos θ  θ + tanθ θ θ + tan θ lim  θ + lim  θ + lim     sin θ  s in θ  θ → 0  sinθ θ → 0 sin θ θ → 0  sin θ 

! = !

1 1 1 = = ! 1 1 1 θ  θ  1 lim + + lim + lim   sin θ lim cos θ θ → 0 sinθ θ → 0 cos θ cos θ  θ → 0  sinθ lim θ →0 θ →0

θ

! !

=

!

46)

lim x →0

1 1 1 + 1 1

=

1 1 . = 2 1+1

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

lim x→0

sin (u ) sin ( x 2 ) = lim . 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 ) u

= lim

u→0

    u sin (u) sin (u ) sin (u) ⋅ = lim  u ⋅  = 0 ⋅1 = 0 .  =  lim u  ⋅  lim  u→0 u  u→0 u  u → 0 u u

47)

1 − tan x sin x − cos x !

lim x → π /4

!

We find that, by a straightforward application of the “limit laws” that this expression gives us a ratio of! π 1 − tan 0 1 − 1 4 = " " , = ! ! 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:

 sin x  1 −    cos x  cos x ⋅ = sin x − cos x cos x

lim

x → π /4

lim

x → π /4

!

=

48)

lim

x →π / 4

− (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 2 −1 −1 = = − 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   .   x →1 x →1 (x −1) (x + 2) x −1   x →1 x + 2  ! x2 + 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 u   1  1 1 ⋅  lim =  lim  = 1⋅ 1+ 2 = 3 .  u → 0 x →1     x + 2 u x +x−2 sin (x −1) 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

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