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Title: Sandwich Theorem
Description: This note is about Sandwich Theorem which will help ypu to solve limits ( which couldn't be done easily).
Description: This note is about Sandwich Theorem which will help ypu to solve limits ( which couldn't be done easily).
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(Section 2
...
6
...
6: THE SQUEEZE (SANDWICH) THEOREM
LEARNING OBJECTIVES
• Understand and be able to rigorously apply the Squeeze (Sandwich) Theorem
when evaluating limits at a point and “long-run” limits at ( ± ) infinity
...
Example 1 below is one of many basic examples where we use the Squeeze
(Sandwich) Theorem to show that lim f x = 0 , where f ( x ) is the product of a
x
0
()
sine or cosine expression and a monomial of even degree
...
Informally,
( Limit Form
0 bounded )
0
...
Prove that lim f x = 0
...
• We first bound cos
• Multiply all three parts by x 2
so that the middle part becomes
f ( x)
...
1 cos
x2
1
x
x 2 cos
1
1
x
(
x
0
x2
(
x
)
0
)
(Section 2
...
6
...
Therefore,
by the Squeeze (Sandwich)
Theorem, the middle part,
f ( x ) , is forced to approach 0,
also
...
x
( x ) = 0 , and lim x
2
lim
0
x
lim x 2 cos
x
0
0
2
= 0 , so
1
= 0 by the Squeeze
x
Theorem
...
0
Therefore, 0
by the Squeeze
(Sandwich) Theorem
The graph of y = x 2 cos
1
, together with the squeezing graphs of y = x 2
x
and y = x 2 , is below
...
)
§
(Section 2
...
6
...
Example 2 (Handling Complications with Signs)
()
1
Let f x = x 3 sin
3
()
...
x
x
0
§ Solution 1 (Using Absolute Value)
• We first bound sin
1
3
1 sin
,
x
which is real for all x 0
...
The inequality
symbols would have to be
reversed for x < 0
...
We could write
0
sin
(
1
)
1
x 0 ,
x
but we assume that absolute
values are nonnegative
...
We know
x3 > 0
(
x
x 3 sin
4
a
3
)
x
0
...
”
• If, say, a
1
4 , then
4
...
6: The Squeeze (Sandwich) Theorem) 2
...
4
• Now, apply the Squeeze
(Sandwich) Theorem
...
Shorthand: As x
x3
0,
1
x 3 sin
0
3
x3
x
(
)
x
0
...
x
Assume x > 0 , since we are taking a limit as x
x
• We first bound sin
0
1
3
3
0+
...
• Multiply all three parts by
x 3 so that the middle part
becomes f ( x )
...
• Now, apply the Squeeze
(Sandwich) Theorem
...
0+ ,
Shorthand: As x
x3
0
x 3 sin
1
3
x
Therefore, 0
by the Squeeze
(Sandwich) Theorem
x3
0
(
)
x>0
...
6: The Squeeze (Sandwich) Theorem) 2
...
5
1
Second, we analyze: lim x 3 sin
...
1
1 sin
,
3
x
which is real for all x 0
...
We know
x3
(
1
x
1
x 3 sin
3
x<0
x3
x
(
)
x<0
)
x 3 < 0 for all x < 0 , so we
reverse the inequality
symbols
...
• Now, apply the Squeeze
(Sandwich) Theorem
...
Shorthand: As x
x3
0 ,
1
x 3 sin
3
0
x3
x
Therefore, 0
by the Squeeze
(Sandwich) Theorem
Now, lim+ x 3 sin
x
0
lim x 3 sin
x
0
1
3
x
1
3
x
= 0
...
(Section 2
...
6
...
x
0
x
0
x
0
§ Solution
Let I = ( 1, 1) \ {0}
...
Shorthand: As x 0 ,
x6
0
x4
Therefore, 0
by the Squeeze
(Sandwich) Theorem
x
I
)
0
WARNING 3: The direction of the
1
confuse students
...
, and
16 4
4
We conclude: lim x 4 = 0
...
We only need it to hold true on some punctured neighborhood of 0 so
that we may apply the Squeeze (Sandwich) Theorem to the two-sided limit
lim x 4
...
”
x
0
As seen below, the graphs of y = x 6 and y = x 2 squeeze (from below and
above, respectively) the graph of y = x 4 on I
...
§
(Section 2
...
6
...
The Squeeze (Sandwich) Theorem
()
f x
(
) , then xlima f ( x ) = L
...
()
()
If lim B x = L and lim T x = L L
x
a
x
a
()
()
T x on a punctured
Variation for Right-Hand Limits at a Point
()
( ) T ( x ) on some right-neighborhood of a
...
x a
x a
a
Let B x
f x
Variation for Left-Hand Limits at a Point
()
Let B x
\
If lim
x
a
( ) T ( x ) on some left-neighborhood of a
...
x a
a
f x
PART C: VARIATIONS FOR “LONG-RUN” LIMITS
In the upcoming Example 4, f ( x ) is the quotient of a sine or cosine expression
and a polynomial
...
Informally,
Limit Form
bounded
±
0
...
6: The Squeeze (Sandwich) Theorem) 2
...
8
Example 4 (Applying the Squeeze (Sandwich) Theorem to a “Long-Run” Limit;
Revisiting Section 2
...
x
§ Solution to a)
Assume x > 0 , since we are taking a limit as x
...
1 sin x 1
(
• Divide all three parts by x
( x > 0 ) so that the middle
part becomes f ( x )
...
sin x
x
x>0
(
1
x
)
x>0
)
lim
1
1
= 0 , and lim = 0 , so
x
x
x
lim
sin x
= 0 by the Squeeze Theorem
...
• We first bound sin x
...
But x < 0 ,
so we must reverse the
inequality symbols
...
1
x
sin x
x
(
x<0
(
1
x
1
x
)
x<0
(
)
x<0
)
)
x>0
...
6: The Squeeze (Sandwich) Theorem) 2
...
9
...
x
x
lim
1
= 0 , and lim
x
x
lim
sin x
= 0 by the Squeeze Theorem
...
We can now justify the HA at y = 0 (the x-axis)
...
) §
Variation for “Long-Run” Limits to the Right
()
Let B x
()
f x
()
()
()
(
If lim B x = L and lim T x = L L
x
(
), c
f ( x) = L
...
• In Example 4a, we used c = 0
...
Variation for “Long-Run” Limits to the Left
()
Let B x
If lim
x
( ) T ( x ) on some x-interval of the form ( , c ) , c
...
x
f x
Title: Sandwich Theorem
Description: This note is about Sandwich Theorem which will help ypu to solve limits ( which couldn't be done easily).
Description: This note is about Sandwich Theorem which will help ypu to solve limits ( which couldn't be done easily).