2.10. The limit of trigonometric functions.
The first remarkable limit
First of all, let us consider principle called the squeeze principle.
The squeeze principle:
If
and
then
Theorem
1: Let
denote the sine of an angle of
radians.
Then
Sometimes this limit is also called ‘the first remarkable limit’.
Theorem
2: Let
denote the cosine of an angle of
radians.
Then
As
or
,
the values of sin x
and cos x oscillate
repeatedly between –1 and 1 without approaching any fixed real
value. Thus, the limits
,
,
,
do not exist. We shall say that they fail to exist due to
oscillation.
Example:
Find
Solution:
Let
,
as x
0,
.
Thus,
=
=
=
.
So,
.
In
particular, if a=2,
then
.
Example:
Find
Solution:
=
=
.
Example:
Find
Solution: Let us divide numerator and denominator by x
=
=
.
Example:
Find
Solution:
=
=
=
=
=11=1.
Example:
Find
Solution:
=
=
=
.
Example:
Find
Solution: As x0 then numerator and denominator approaches zero. Let us multiply numerator and denominator by the conjugate of denominator:
=
=
=
=1(3+3)=6.
Example:
Find
Solution:
Observe that as x2,
we shall have
.
Let
.
We obtain
=
=
=
=
=
=
=
=
.
Exercises
In exercises 1-18 find the limits.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19. Find a nonzero value for the constant k so that
will
be continuous at x=0.
Answers
1. 3; 2. 0; 3. 7/3; 4. 1; 5. 2; 6. –25/49; 7. 3; 8. 1; 9. a/b; 10. 1/8;
11. 1/3; 12. 4; 13. –1/2; 14. 3; 15. 1/36 ; 16. 3/2; 17. –1; 18. /4;
19. 1/2.
2.11. The number e. Second remarkable limit
Number e is the limit
(1) 
or
(2) 
Limits (1) and (2) are equivalent and called the second remarkable limits.
To evaluate
there are following possible cases.
a) If
and
then C=AB
b) If
and
then we apply
or
c) If
and
then we assume
,
where
as x
a
and
=
=
.
Example:
Find
Solution:
As
,
expression
and we get indeterminate form
.
Let us introduce
by
.
If
then
.
Thus,
=
=
Using (2) we obtain
=
=
(3) = ;
In
particular, if k=3,
then
=
Example:
Find
Solution:
Since
Using (2) we obtain
=
=
=1.
Example:
Find
Solution:
Let us divide numerator and denominator by x, and then use (3)
=
=
.
Example:
Find
Solution:
=
Let
.
Then
.
As
,
then
.
We obtain
=
=
=
=
.
Example:
Find
Solution:
=
Let
.
Then
=
=
=
=
.
Exercises
In exercises 1-12 find the limits.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Answers
1.
;
2.
;
3.
4; 4.
1; 5.
;
6.
;
7.
;
8.
;
9.
;
10.
2; 11.
4/5; 12.
.