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Calculus BC · Unit 1 · 1.8
AP Calculus · Limits and Continuity
Determining Limits Using the Squeeze Theorem
Trap the function between two curves that meet. If they meet, it has nowhere else to go.
key idea$g\le f\le h\ \text{and}\ \lim g=\lim h=L\ \Rightarrow\ \lim f=L$
The theorem
The Squeeze Theorem
Hypotheses — all must hold
- $g(x)\le f(x)\le h(x)$ for all $x$ near $c$ (the point $c$ itself need not be included).
- $\lim_{x\to c}g(x)=L$ and $\lim_{x\to c}h(x)=L$ — the same $L$.
What you may then conclude
$\lim_{x\to c}f(x)=L$. The middle function has nowhere else to go.
Bounds that do not meet prove nothing
In the right-hand panel, $-1\le\sin\!\left(\tfrac1x\right)\le1$ holds perfectly well, but the bounds converge to $-1$ and $1$ — different values. The trap never closes, and $\lim_{x\to0}\sin\!\left(\tfrac1x\right)$ genuinely does not exist. Being bounded is not enough.
the trap closing at $x=0$
Checking the hypotheses
Almost all the credit is in showing the two conditions hold. Write the inequality chain first, then the two limits, then the conclusion — in that order, every time.
The three-line proof
- Bound the oscillating factor. Start from a fact that is always true, such as $-1\le\sin\theta\le1$ or $0\le\cos^{2}\theta\le1$.
- Multiply through by the decaying factor to build the chain around $f$. Multiplying by a negative quantity reverses the inequalities — handle $x\lt 0$ separately or use $\lvert x\rvert$.
- Take the limit of both bounds and show they agree. Then state the conclusion by the Squeeze Theorem, naming it.
| Signal in the problem | Why squeeze is the tool | The bound to start from |
|---|
| $\sin$ or $\cos$ of something that blows up | the factor has no limit but is bounded | $-1\le\sin\theta\le1$ |
| a bounded factor times something $\to0$ | the decay kills the oscillation | multiply the bound by the decaying factor |
| $\lfloor x\rfloor$ or a rounding function | bounded difference from $x$ | $x-1\lt\lfloor x\rfloor\le x$ |
| limit laws give $0\times(\text{no limit})$ | the product law does not apply | bound the offending factor instead |
The tell is a factor that oscillates without settling multiplied by one that decays to zero. The limit laws cannot handle that product; the squeeze can.
Worked examples
EX 1
The standard case
$x^{2}\sin\!\left(\tfrac1x\right)$Find $\lim\limits_{x\to0}x^{2}\sin\!\left(\dfrac{1}{x}\right)$.
1
Note why the laws fail
As $x\to0$, $\sin\!\left(\tfrac1x\right)$ has no limit, so the product law cannot be used.
2
Bound the oscillating factor
For every $x\neq0$: $-1\le\sin\!\left(\tfrac1x\right)\le1$.
3
Multiply by $x^{2}$
$x^{2}\ge0$ always, so the inequalities keep their direction: $-x^{2}\le x^{2}\sin\!\left(\tfrac1x\right)\le x^{2}$.
4
Take both outer limits
$\lim_{x\to0}\left(-x^{2}\right)=0$ and $\lim_{x\to0}x^{2}=0$ — the same value.
5
Conclude
By the Squeeze Theorem, $\lim_{x\to0}x^{2}\sin\!\left(\tfrac1x\right)=0$.
EX 2
When the squeeze does not apply
$\sin\!\left(\tfrac1x\right)$Can the Squeeze Theorem show that $\lim\limits_{x\to0}\sin\!\left(\dfrac{1}{x}\right)$ exists?
1
Write the natural bounds
$-1\le\sin\!\left(\tfrac1x\right)\le1$ — hypothesis 1 holds.
2
Test hypothesis 2
$\lim_{x\to0}(-1)=-1$ and $\lim_{x\to0}(1)=1$. These are different.
3
Conclude about the theorem
Hypothesis 2 fails, so the Squeeze Theorem says nothing. It does not conclude the limit is $0$ or anything else.
4
Determine the truth separately
The function takes every value in $[-1,1]$ infinitely often near $0$, so the limit genuinely does not exist.
5
The lesson
Bounded is not enough. The two bounds must converge to the same number.
EX 3
Careful with negative multipliers
$x\cos\!\left(\tfrac1x\right)$Find $\lim\limits_{x\to0}x\cos\!\left(\dfrac{1}{x}\right)$.
1
Bound the cosine
$-1\le\cos\!\left(\tfrac1x\right)\le1$.
2
Spot the sign problem
Multiplying by $x$ reverses the chain when $x\lt 0$, so a single chain in $x$ is not valid on both sides.
3
Use absolute values instead
$\left\lvert x\cos\!\left(\tfrac1x\right)\right\rvert\le\lvert x\rvert$, hence $-\lvert x\rvert\le x\cos\!\left(\tfrac1x\right)\le\lvert x\rvert$ for all $x$.
4
Take the outer limits
$\lim_{x\to0}\left(-\lvert x\rvert\right)=0=\lim_{x\to0}\lvert x\rvert$.
5
Conclude
By the Squeeze Theorem the limit is $0$. Wrapping in $\lvert\cdot\rvert$ sidesteps the sign issue entirely.
EX 4
Squeeze at infinity
$\dfrac{\sin x}{x}$Find $\lim\limits_{x\to\infty}\dfrac{\sin x}{x}$.
1
Bound the numerator
$-1\le\sin x\le1$ for all $x$.
2
Divide by $x$
For $x\gt 0$ the direction is preserved: $-\dfrac1x\le\dfrac{\sin x}{x}\le\dfrac1x$.
3
Take the outer limits
$\lim_{x\to\infty}\left(-\tfrac1x\right)=0$ and $\lim_{x\to\infty}\tfrac1x=0$.
4
Conclude
By the Squeeze Theorem, $\lim_{x\to\infty}\dfrac{\sin x}{x}=0$.
5
Distinguish it
This is not the famous $\lim_{x\to0}\tfrac{\sin x}{x}=1$. Same expression, different approach point, different answer — and a different method.
When it fails
The bounds must meet
Converging to different values proves nothing at all. That is the whole content of Example 2.
Multiplying by a negative flips the chain
$a\le b\le c$ becomes $-a\ge-b\ge-c$. Use $\lvert x\rvert$ bounds to keep one chain valid on both sides.
The inequality must hold near $c$, not just at it
A bound true only at $x=c$ says nothing about the approach. It must hold on an interval around $c$ (the point itself excepted).
Name the theorem
“The limit is $0$” without citing the Squeeze Theorem and showing both bounds is an incomplete justification.