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Wize University Calculus 1 Textbook > Limits

The Squeeze Theorem

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The Squeeze Theorem

Some limits can be solved by bounding a function above and below. This lets us squeeze the function between two limits which turn out to be the same when done correctly!


Squeeze Theorem (Squeeze Law)

Letf(x)≤g(x)≤h(x)f(x)\leq g(x)\leq h(x) , for all xx around a,a, except possibly at x=a.x=a.

if lim⁡x→af(x)=lim⁡x→ah(x)=L\text{if }\displaystyle\lim_{x\rightarrow a}f(x)=\displaystyle\lim_{x\rightarrow a}h(x)=L \\

then lim⁡x→ag(x)=L.\text{then }\displaystyle\lim_{x\rightarrow a}g(x)=L.




Wize Tip
Notice how the f(x) and h(x) functions bound the g(x) function above and below and then have the same limit!


Watch Out!
The Squeeze Theorem is usually only applied on functions with well known boundaries such as Sinx, Cosx, and Arctanx.

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Example: The Squeeze Theorem

Find the following limit

lim⁡x→0+(xsin⁡1x2+x)\displaystyle \lim_{x\rightarrow 0^+}\left(x\sin{\frac{1}{x^2+x}}\right)


−1≤sin⁡(1x2+x)≤1−x≤x⋅sin⁡(1x2+x)≤xlim⁡x→0+(−x)=0,lim⁡x→0+x=0∴0≤lim⁡x→0+x⋅sin⁡(1x2+x)≤0⇒lim⁡x→0+x⋅sin⁡(1x2+x)=0\begin{array}{c} -1 \leq \sin\left(\dfrac{1}{x^2+x}\right) \leq 1\\ -x \leq x\cdot\sin\left(\dfrac{1}{x^2+x}\right) \leq x\\ \displaystyle\lim_{x \to 0^+}(-x) = 0 \quad , \quad \lim_{x \to 0^+}x = 0\\ \displaystyle\therefore 0 \leq \lim_{x \to 0^+}x\cdot\sin\left(\dfrac{1}{x^2+x}\right) \leq 0\\ \Rightarrow {\fbox{$ \displaystyle\lim_{x \to 0^+}x\cdot\sin\left(\dfrac{1}{x^2+x}\right)=0$}}\end{array}


Find the following limit

lim⁡x→0x2arctan⁡1x\displaystyle \lim_{x\rightarrow0}x^2\arctan{\frac{1}{x}}




Extra Practice
The Squeeze Theorem
Use the Squeeze Theorem to determine the limit lim⁡x→01+2xsin⁡1xx2+1\lim\limits_{x\rightarrow 0}\frac{1+2x\sin\frac{1}{x}}{x^2+1}
The Squeeze Theorem
Let f(x)f(x) be a function that satisfies
x−1x2−1≤f(x)≤12x2−x+1\frac{x-1}{x^2-1}\leq f(x)\leq \frac{1}{2x^2-x+1}
Only one of the statements below doesn’t follow from the Squeeze Theorem. Which one is it?
The Squeeze Theorem
Evaluate lim⁡x→∞ e1x⋅cos⁡(x2−2)ln⁡x\displaystyle\lim_{x\to\infty}\ \frac{e^{\frac{1}{x}}\cdot\cos\left(x^2-2\right)}{\ln x}.
The Squeeze Theorem
Suppose that f(x) is a function such that
3x2−x+6≤f(x)≤Ax−33x^2-x+6 \leq f(x) \leq Ax-3
for x close to 4. In order for us to use the Squeeze Theorem to find lim⁡x→4f(x)\lim\limits_{x\to4}f(x), what is the value of A and the limit lim⁡x→4f(x)\lim\limits_{x\to4}f(x)?
The Squeeze Theorem
Use the Squeeze Theorem to determine the limit: lim⁡x→01+2xcos⁡1xx2+1\displaystyle\lim_{x\to0}\frac{1+2x\cos\frac{1}{x}}{x^2+1}
The Squeeze Theorem
Compute lim⁡x →0+ x⋅2sin⁡(1x)\displaystyle \lim_{x\ \rightarrow0^+}\ x\cdot2^{\sin\left(\frac{1}{x}\right)} if it exists.
The Squeeze Theorem
Practice: Squeeze Theorem
If e2ln⁡x−1≤f(x)≤x2−x+2e^{2\ln x}-1\le f\left(x\right)\le x^2-x+2 on the interval [−1,4]\left[-1,4\right], determine the lim⁡x→3f(x)\displaystyle\lim_{x\to3}f\left(x\right).
The Squeeze Theorem
Evaluate lim⁡x→∞ e1x⋅cos⁡(x2−2)ln⁡x\displaystyle\lim_{x\to\infty}\ \frac{e^{\frac{1}{x}}\cdot\cos\left(x^2-2\right)}{\ln x}.
The squeeze theorem
Let 3x2−2x+14x2−5≤f(x)≤34e1x\frac{3x^2-2x+1}{4x^2-5}\le f\left(x\right)\le\frac{3}{4}e^{\frac{1}{x}} for all xx values. Then lim⁡x→∞f(x)\displaystyle \lim_{x\to\infty}f\left(x\right) is
The Squeeze Theorem
Evaluate
lim⁡x→0x2sin⁡(1x2)\lim\limits_{x\to 0} x^2 \sin\left(\frac{1}{x^2}\right)
lim⁡x→∞xsin⁡xx2+1\displaystyle\lim_{x\rightarrow\infty}\dfrac{x\sin{x}}{x^2+1}
Let f(x)f(x) be a function that satisfies: 2−xx2−4 ≤ f(x) ≤ 12x2−3x+2\dfrac{2-x}{x^2-4}\ \le\ f\left(x\right)\ \le\ \dfrac{1}{2x^2-3x+2} for x≤0x\le0 and x≥2x\ge2. Only one of the statements below doesn't follow from the squeeze Theorem. Which one is it ?
Q.\textbf{Q.} Find lim⁡x→0xsin⁡(1x2+x)\displaystyle \lim_{x\to 0}x\sin{\left(\frac{1}{x^2+x}\right)}
The Squeeze Theorem
Use the Squeeze Theorem to determine the limit: lim⁡x→01+2xcos⁡1xx2+1\displaystyle\lim_{x\to0}\frac{1+2x\cos\frac{1}{x}}{x^2+1}
The Squeeze Theorem
Suppose that f(x) is a function such that
3x2−x+6≤f(x)≤Ax−33x^2-x+6 \leq f(x) \leq Ax-3
for x close to 4. In order for us to use the Squeeze Theorem to find lim⁡x→4f(x)\lim\limits_{x\to4}f(x), what is the value of A and the limit lim⁡x→4f(x)\lim\limits_{x\to4}f(x)?
Practice: Squeeze Theorem
Evaluate lim⁡x→∞(cos⁡ xx+tan⁡−1x)\lim\limits_{x\rightarrow\infin}\left(\frac{ \cos\ x}{\sqrt{x}}+\tan^{-1}x\right).
Use the Squeeze Theorem to determine the value of lim⁡x→01+2xsin⁡1xx2+1\lim\limits_{x\rightarrow 0}\frac{1+2x\sin\frac{1}{x}}{x^2+1}.
🌶️ TOUGH! Let f(x)f(x) be a function that satisfies
x−1x2−1≤f(x)≤12x2−x+1\frac{x-1}{x^2-1}\leq f(x)\leq \frac{1}{2x^2-x+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→01+2xsin⁡1xx2+1\displaystyle\lim_{x\to0}\frac{1+2x\sin\frac{1}{x}}{x^2+1}
The Squeeze Theorem
Use the Squeeze Theorem to determine the limit lim⁡x→01+2xsin⁡1xx2+1\lim\limits_{x\rightarrow 0}\frac{1+2x\sin\frac{1}{x}}{x^2+1}
The Squeeze Theorem
Let f(x)f(x) be a function that satisfies
x−1x2−1≤f(x)≤12x2−x+1\frac{x-1}{x^2-1}\leq f(x)\leq \frac{1}{2x^2-x+1}
Only one of the statements below doesn’t follow from the Squeeze Theorem. Which one is it?
🌶️ TOUGH! Let f(x)f(x) be a function that satisfies:
x−2x2−4≤ f(x)≤12x2−3x+2\frac{x-2}{x^{2}-4}\le\ f\left(x\right)\le\frac{1}{2x^{2}-3x+2}
forx≤0x\le0 and x≥2x\ge2. Only one of the statements below doesn't follow from the Squeeze Theorem. Which one is it ?
Determine lim⁡x→∞e−x⋅cos⁡(ex)\displaystyle\lim_{x\to\infty}e^{-x}\cdot \cos\left(e^x\right).
The Squeeze Theorem
Compute lim⁡x →0+ x⋅2sin⁡(1x)\displaystyle \lim_{x\ \rightarrow0^+}\ x\cdot2^{\sin\left(\frac{1}{x}\right)} if it exists.
The Squeeze Theorem
lim⁡x→∞(cos⁡ xx+tan⁡−1x)\lim\limits_{x\rightarrow\infin}\left(\frac{ \cos\ x}{\sqrt{x}}+\tan^{-1}x\right)
The Squeeze Theorem
Practice: Squeeze Theorem
If e2ln⁡x−1≤f(x)≤x2−x+2e^{2\ln x}-1\le f\left(x\right)\le x^2-x+2 on the interval [−1,4]\left[-1,4\right], determine the lim⁡x→3f(x)\displaystyle\lim_{x\to3}f\left(x\right).
The Squeeze Theorem
Evaluate lim⁡x→∞ e1x⋅cos⁡(x2−2)ln⁡x\displaystyle\lim_{x\to\infty}\ \frac{e^{\frac{1}{x}}\cdot\cos\left(x^2-2\right)}{\ln x}.
Limit Techniques: The Squeeze Theorem
lim⁡x→0xarctan⁡1x\displaystyle \lim_{x\rightarrow0}x\arctan{\frac{1}{x}}