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🌶️ TOUGH! Letf(x) be a function that satisfies: (x-2)/(x^2-4)≤ f(x)≤1/(2x^2-…
Related Topics
Wize University Calculus 1 Textbook > Limits
The Squeeze Theorem
3 Activities
🌶️
TOUGH!
Let
f
(
x
)
f(x)
f
(
x
)
be a function that satisfies:
x
−
2
x
2
−
4
≤
f
(
x
)
≤
1
2
x
2
−
3
x
+
2
\frac{x-2}{x^{2}-4}\le\ f\left(x\right)\le\frac{1}{2x^{2}-3x+2}
x
2
−
4
x
−
2
≤
f
(
x
)
≤
2
x
2
−
3
x
+
2
1
for
x
≤
0
x\le0
x
≤
0
and
x
≥
2
x\ge2
x
≥
2
. Only one of the statements below doesn't follow from the Squeeze Theorem. Which one is it ?
lim
x
→
−
∞
f
(
x
)
=
0
\lim\limits_{x\rightarrow-\infin}f(x)=0
x
→
−
∞
lim
f
(
x
)
=
0
lim
x
→
−
2
f
(
x
)
=
1
16
\lim\limits_{x\rightarrow-2}f(x)=\frac{1}{16}
x
→
−
2
lim
f
(
x
)
=
16
1
lim
x
→
0
−
f
(
x
)
=
1
2
\lim\limits_{x\rightarrow0^-}f(x)=\frac{1}{2}
x
→
0
−
lim
f
(
x
)
=
2
1
lim
x
→
2
+
f
(
x
)
=
1
4
\lim\limits_{x\rightarrow2^+}f(x)=\frac{1}{4}
x
→
2
+
lim
f
(
x
)
=
4
1
lim
x
→
∞
f
(
x
)
=
0
\lim\limits_{x\rightarrow\infin}f(x)=0
x
→
∞
lim
f
(
x
)
=
0
I don't know
Check Submission
More The Squeeze Theorem Questions:
The Squeeze Theorem
Use the Squeeze Theorem to determine the limit
lim
x
→
0
1
+
2
x
sin
1
x
x
2
+
1
\lim\limits_{x\rightarrow 0}\frac{1+2x\sin\frac{1}{x}}{x^2+1}
x
→
0
lim
x
2
+
1
1
+
2
x
s
i
n
x
1
The Squeeze Theorem
Let
f
(
x
)
f(x)
f
(
x
)
be a function that satisfies
x
−
1
x
2
−
1
≤
f
(
x
)
≤
1
2
x
2
−
x
+
1
\frac{x-1}{x^2-1}\leq f(x)\leq \frac{1}{2x^2-x+1}
x
2
−
1
x
−
1
≤
f
(
x
)
≤
2
x
2
−
x
+
1
1
Only one of the statements below doesn’t follow from the Squeeze Theorem. Which one is it?
The Squeeze Theorem
Evaluate
lim
x
→
∞
e
1
x
⋅
cos
(
x
2
−
2
)
ln
x
\displaystyle\lim_{x\to\infty}\ \frac{e^{\frac{1}{x}}\cdot\cos\left(x^2-2\right)}{\ln x}
x
→
∞
lim
ln
x
e
x
1
⋅
cos
(
x
2
−
2
)
.
The Squeeze Theorem
Suppose that
f
(
x
) is a function such that
3
x
2
−
x
+
6
≤
f
(
x
)
≤
A
x
−
3
3x^2-x+6 \leq f(x) \leq Ax-3
3
x
2
−
x
+
6
≤
f
(
x
)
≤
A
x
−
3
for
x
close to 4. In order for us to use the Squeeze Theorem to find
lim
x
→
4
f
(
x
)
\lim\limits_{x\to4}f(x)
x
→
4
lim
f
(
x
)
, what is the value of
A
and the limit
lim
x
→
4
f
(
x
)
\lim\limits_{x\to4}f(x)
x
→
4
lim
f
(
x
)
?
The Squeeze Theorem
Use the Squeeze Theorem to determine the limit:
lim
x
→
0
1
+
2
x
cos
1
x
x
2
+
1
\displaystyle\lim_{x\to0}\frac{1+2x\cos\frac{1}{x}}{x^2+1}
x
→
0
lim
x
2
+
1
1
+
2
x
cos
x
1
The Squeeze Theorem
Compute
lim
x
→
0
+
x
⋅
2
sin
(
1
x
)
\displaystyle \lim_{x\ \rightarrow0^+}\ x\cdot2^{\sin\left(\frac{1}{x}\right)}
x
→
0
+
lim
x
⋅
2
s
i
n
(
x
1
)
if it exists.
The Squeeze Theorem
Practice: Squeeze Theorem
If
e
2
ln
x
−
1
≤
f
(
x
)
≤
x
2
−
x
+
2
e^{2\ln x}-1\le f\left(x\right)\le x^2-x+2
e
2
l
n
x
−
1
≤
f
(
x
)
≤
x
2
−
x
+
2
on the interval
[
−
1
,
4
]
\left[-1,4\right]
[
−
1
,
4
]
, determine the
lim
x
→
3
f
(
x
)
\displaystyle\lim_{x\to3}f\left(x\right)
x
→
3
lim
f
(
x
)
.
The Squeeze Theorem
Evaluate
lim
x
→
∞
e
1
x
⋅
cos
(
x
2
−
2
)
ln
x
\displaystyle\lim_{x\to\infty}\ \frac{e^{\frac{1}{x}}\cdot\cos\left(x^2-2\right)}{\ln x}
x
→
∞
lim
ln
x
e
x
1
⋅
cos
(
x
2
−
2
)
.
The squeeze theorem
Let
3
x
2
−
2
x
+
1
4
x
2
−
5
≤
f
(
x
)
≤
3
4
e
1
x
\frac{3x^2-2x+1}{4x^2-5}\le f\left(x\right)\le\frac{3}{4}e^{\frac{1}{x}}
4
x
2
−
5
3
x
2
−
2
x
+
1
≤
f
(
x
)
≤
4
3
e
x
1
for all
x
x
x
values. Then
lim
x
→
∞
f
(
x
)
\displaystyle \lim_{x\to\infty}f\left(x\right)
x
→
∞
lim
f
(
x
)
is
The Squeeze Theorem
Evaluate
lim
x
→
0
x
2
sin
(
1
x
2
)
\lim\limits_{x\to 0} x^2 \sin\left(\frac{1}{x^2}\right)
x
→
0
lim
x
2
sin
(
x
2
1
)
lim
x
→
∞
x
sin
x
x
2
+
1
\displaystyle\lim_{x\rightarrow\infty}\dfrac{x\sin{x}}{x^2+1}
x
→
∞
lim
x
2
+
1
x
sin
x
Let
f
(
x
)
f(x)
f
(
x
)
be a function that satisfies:
2
−
x
x
2
−
4
≤
f
(
x
)
≤
1
2
x
2
−
3
x
+
2
\dfrac{2-x}{x^2-4}\ \le\ f\left(x\right)\ \le\ \dfrac{1}{2x^2-3x+2}
x
2
−
4
2
−
x
≤
f
(
x
)
≤
2
x
2
−
3
x
+
2
1
for
x
≤
0
x\le0
x
≤
0
and
x
≥
2
x\ge2
x
≥
2
. Only one of the statements below doesn't follow from the squeeze Theorem. Which one is it ?
Q.
\textbf{Q.}
Q.
Find
lim
x
→
0
x
sin
(
1
x
2
+
x
)
\displaystyle \lim_{x\to 0}x\sin{\left(\frac{1}{x^2+x}\right)}
x
→
0
lim
x
sin
(
x
2
+
x
1
)
The Squeeze Theorem
Use the Squeeze Theorem to determine the limit:
lim
x
→
0
1
+
2
x
cos
1
x
x
2
+
1
\displaystyle\lim_{x\to0}\frac{1+2x\cos\frac{1}{x}}{x^2+1}
x
→
0
lim
x
2
+
1
1
+
2
x
cos
x
1
The Squeeze Theorem
Suppose that
f
(
x
) is a function such that
3
x
2
−
x
+
6
≤
f
(
x
)
≤
A
x
−
3
3x^2-x+6 \leq f(x) \leq Ax-3
3
x
2
−
x
+
6
≤
f
(
x
)
≤
A
x
−
3
for
x
close to 4. In order for us to use the Squeeze Theorem to find
lim
x
→
4
f
(
x
)
\lim\limits_{x\to4}f(x)
x
→
4
lim
f
(
x
)
, what is the value of
A
and the limit
lim
x
→
4
f
(
x
)
\lim\limits_{x\to4}f(x)
x
→
4
lim
f
(
x
)
?
Practice: Squeeze Theorem
Evaluate
lim
x
→
∞
(
cos
x
x
+
tan
−
1
x
)
\lim\limits_{x\rightarrow\infin}\left(\frac{ \cos\ x}{\sqrt{x}}+\tan^{-1}x\right)
x
→
∞
lim
(
x
c
o
s
x
+
tan
−
1
x
)
.
Use the Squeeze Theorem to determine the value of
lim
x
→
0
1
+
2
x
sin
1
x
x
2
+
1
\lim\limits_{x\rightarrow 0}\frac{1+2x\sin\frac{1}{x}}{x^2+1}
x
→
0
lim
x
2
+
1
1
+
2
x
s
i
n
x
1
.
🌶️
TOUGH!
Let
f
(
x
)
f(x)
f
(
x
)
be a function that satisfies
x
−
1
x
2
−
1
≤
f
(
x
)
≤
1
2
x
2
−
x
+
1
\frac{x-1}{x^2-1}\leq f(x)\leq \frac{1}{2x^2-x+1}
x
2
−
1
x
−
1
≤
f
(
x
)
≤
2
x
2
−
x
+
1
1
Only one of the statements below doesn’t follow from the Squeeze Theorem. Which one is it?
Use the Squeeze Theorem to determine the limit:
lim
x
→
0
1
+
2
x
sin
1
x
x
2
+
1
\displaystyle\lim_{x\to0}\frac{1+2x\sin\frac{1}{x}}{x^2+1}
x
→
0
lim
x
2
+
1
1
+
2
x
sin
x
1
The Squeeze Theorem
Use the Squeeze Theorem to determine the limit
lim
x
→
0
1
+
2
x
sin
1
x
x
2
+
1
\lim\limits_{x\rightarrow 0}\frac{1+2x\sin\frac{1}{x}}{x^2+1}
x
→
0
lim
x
2
+
1
1
+
2
x
s
i
n
x
1
The Squeeze Theorem
Let
f
(
x
)
f(x)
f
(
x
)
be a function that satisfies
x
−
1
x
2
−
1
≤
f
(
x
)
≤
1
2
x
2
−
x
+
1
\frac{x-1}{x^2-1}\leq f(x)\leq \frac{1}{2x^2-x+1}
x
2
−
1
x
−
1
≤
f
(
x
)
≤
2
x
2
−
x
+
1
1
Only one of the statements below doesn’t follow from the Squeeze Theorem. Which one is it?
Determine
lim
x
→
∞
e
−
x
⋅
cos
(
e
x
)
\displaystyle\lim_{x\to\infty}e^{-x}\cdot \cos\left(e^x\right)
x
→
∞
lim
e
−
x
⋅
cos
(
e
x
)
.
The Squeeze Theorem
Compute
lim
x
→
0
+
x
⋅
2
sin
(
1
x
)
\displaystyle \lim_{x\ \rightarrow0^+}\ x\cdot2^{\sin\left(\frac{1}{x}\right)}
x
→
0
+
lim
x
⋅
2
s
i
n
(
x
1
)
if it exists.
The Squeeze Theorem
lim
x
→
∞
(
cos
x
x
+
tan
−
1
x
)
\lim\limits_{x\rightarrow\infin}\left(\frac{ \cos\ x}{\sqrt{x}}+\tan^{-1}x\right)
x
→
∞
lim
(
x
c
o
s
x
+
tan
−
1
x
)
The Squeeze Theorem
Practice: Squeeze Theorem
If
e
2
ln
x
−
1
≤
f
(
x
)
≤
x
2
−
x
+
2
e^{2\ln x}-1\le f\left(x\right)\le x^2-x+2
e
2
l
n
x
−
1
≤
f
(
x
)
≤
x
2
−
x
+
2
on the interval
[
−
1
,
4
]
\left[-1,4\right]
[
−
1
,
4
]
, determine the
lim
x
→
3
f
(
x
)
\displaystyle\lim_{x\to3}f\left(x\right)
x
→
3
lim
f
(
x
)
.
The Squeeze Theorem
Evaluate
lim
x
→
∞
e
1
x
⋅
cos
(
x
2
−
2
)
ln
x
\displaystyle\lim_{x\to\infty}\ \frac{e^{\frac{1}{x}}\cdot\cos\left(x^2-2\right)}{\ln x}
x
→
∞
lim
ln
x
e
x
1
⋅
cos
(
x
2
−
2
)
.
Limit Techniques: The Squeeze Theorem
lim
x
→
0
x
arctan
1
x
\displaystyle \lim_{x\rightarrow0}x\arctan{\frac{1}{x}}
x
→
0
lim
x
arctan
x
1