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Wize AP Calculus (AB) Textbook > Limits & Continuity

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}}