Download MATH 1910 - Trigonometric Limits and the Squeeze

MATH 1910 - Trigonometric Limits and the Squeeze Theorem
 Finding limits involving trigonometric functions.
Two important limits involving trigonometric functions are
x
2. lim 1 − xcos x  0
1. lim sin
x 1
x→0
x→0
The first one we have look at earlier. It can be proved using geometry. Let’s derive the second one
from the first.
x
Our goal will be to transform 1 − xcos x into something involving the expression sin
x . This process
involves making use of trigonometric identities & other algebraic properties (this is one reason why
you spent so much time in precalculus verifying trigonometric identities).
2
sin 2 x
lim 1 − xcos x  lim 1 − xcos x  1  cos x  lim 1 − cos x  lim
x→0
x→0
x→0 x1  cos x
x→0 x1  cos x
1  cos x
x
sin x  lim
sin x
sin x
 lim sin
x  1  cos x lim
x
x→0
x→0
x→0
1  cos x
This is 1
 1
sin 0
 1 0  10  0
1  cos 0
11
Here’s some more examples:
x
1. Find lim tan
x
x→0
Notice that direct substitution does not work. We try and obtain one of the limits stated above.
x
lim tan
x  lim
x→0
x→0
sin x
cos x
sin x 1
sin x  1
x  lim
x
cos x
cos x  x  lim
x→0
x→0
x  lim 1
 lim sin
 1 1  1 1  1
x
x→0
x→0 cos x
cos 0
1
2. Find lim sinx4x
x→0
x
This looks almost like sin
x . Notice that
sin 4x  sin 4x  4  4 sin 4x
x
x
4x
4
Now, if we let y  4x, we have
sin y
4 sin 4x  4
y
4x
Now since y  4x, we have x → 0 if and only if y → 0. Thus we have
sin y
sin y
lim sinx4x  lim 4 sin 4x  lim 4
 4 lim y
y
x→0
x→0
y→0
y→0
4x
Check this numerically and graphically.
3. Find lim sin 3 sin 2
→0
 sin 5
 41  4
1
lim sin 3 sin 2  lim sin 3  sin 2
→0
→0

sin 5
 sin 5
sin 2/
 lim 3  sin 3  2  5 
→0

2 5 sin 5/
3
sin 2/2
 lim 3  sin 3  2  1 
→0
5
3
1
sin 5/5
sin 2/2
 lim 6  sin 3 
→0
5
3
sin 5/5
sin 2/2
 6 lim sin 3 
5 →0
3
sin 5/5
sin 2/2
 6 lim sin 3  lim
→0 sin 5/5
5 →0 3
 6 1 1  6
5
5
1
 The Squeeze (or Sandwich) Theorem
If fx ≤ hx ≤ gx for all x near a (except possibly at a) and
fx  L  lim
gx
lim
x→a
x→a
then
hx  L
lim
x→a
In essence, this theorem says that the function h is squeezed (sandwiched) between the functions f
and g as x approaches a, forcing h to have the same limit as f & g.
Example:
Find lim x 2 sin 1x . Note that you cannot just use direct substitution since lim sin 1x DNE. We try to
x→0
find to functions that bound x 2 sin
for both functions. Note that
1
x
x→0
and for which the limit as x approaches 0 exists and is the same
− 1 ≤ sin 1x ≤ 1 for all x ≠ 0
Since x 2  0, we have
− x 2 ≤ x 2 sin 1x ≤ x 2 for all x
But note that lim−x 2   0  lim x 2 . Thus, by the Squeeze Theorem,
x→0
x→0
lim x 2 sin 1x  0
x→0
2