Determining Limits Using the Squeeze Theorem
Introduced| English |
|---|
| Squeeze Theorem/skwiːz ˈθɪərəm/ |
| inequality/ɪniːˈkwɒlɪti/ |
Trap a wild function between two tame ones
- Some limits resist substitution and algebra — like $\displaystyle\lim_{x\to0}x^2\sin\tfrac1x$, which wiggles infinitely fast.
- The escape: trap the messy function between two friendlier ones.
- If the upper and lower "walls" both head to the same height, the trapped function has nowhere else to go.
- This is the Squeeze Theorem 夹逼定理 (also called the Sandwich Theorem).
Trapping a function between two others with equal limits uses the ____ Theorem.
Also called the Sandwich Theorem.
What the theorem says
- Suppose near $c$ (except maybe at $c$ itself) we have $g(x)\le f(x)\le h(x)$.
- If $\displaystyle\lim_{x\to c}g(x)=\lim_{x\to c}h(x)=L$, then $\displaystyle\lim_{x\to c}f(x)=L$ too.
- The two bounds squeeze $f$ onto the single value $L$.
- You need two things: a valid inequality 不等式 near $c$, and equal limits for the bounds.
To apply the Squeeze Theorem to $f$ near $c$, you need all of...
You need two-sided bounds with equal limits, valid near $c$. $f$ need not be a polynomial — that is the whole point.
The classic wiggle
- $\sin\tfrac1x$ swings between $-1$ and $1$ forever as $x\to0$ — no limit on its own.
- But multiply by $x^2$: since $-1\le\sin\tfrac1x\le1$, we get $-x^2\le x^2\sin\tfrac1x\le x^2$.
- Both walls $\pm x^2 \to 0$ as $x\to0$.
- Squeezed between them, $\displaystyle\lim_{x\to0}x^2\sin\tfrac1x=0$.
The squeezing wall $y=x^2$
y = ax²
The parabola $y=x^2$ and its mirror $-x^2$ both pinch to $0$ at the origin — anything trapped between them is forced to $0$.
Using $-x^2\le x^2\sin\tfrac1x\le x^2$, find $\displaystyle\lim_{x\to0}x^2\sin\tfrac1x$.
Both walls $\pm x^2\to 0$, so the trapped function $\to 0$.
Checking the setup honestly
- The inequality only has to hold near $c$, not everywhere — a small interval around $c$ is enough.
- The bounds' limits must be equal; if the walls head to different heights, the theorem says nothing.
- Squeeze is the standard route to the famous $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$, from the geometric bounds $\cos x \le \frac{\sin x}{x}\le 1$.
- Always name your two bounding functions explicitly — that is the heart of the argument.
Bounding $f$ from above only, with $f(x)\le h(x)$, is enough to apply the Squeeze Theorem.
You need bounds on both sides with equal limits — one bound proves nothing.
The Squeeze Theorem, with $\cos x\le\tfrac{\sin x}{x}\le 1$, gives $\displaystyle\lim_{x\to0}\tfrac{\sin x}{x}=$
As $x\to0$, $\cos x\to1$ and the upper bound is $1$, so the middle is squeezed to $1$.
If the lower bound heads to $2$ and the upper bound heads to $5$, the Squeeze Theorem tells you the trapped limit is...
The bounds must have equal limits; unequal walls give no conclusion.
The Squeeze Theorem needs the bounds to squeeze from both sides with equal limits. If you only bound $f$ from above ($f\le h$), or the two bounds head to different values, you have proved nothing. And the inequality must genuinely hold near $c$ — check it, don't assume it.
Show $\displaystyle\lim_{x\to0}x^2\cos\tfrac1x=0$.
- For all $x\neq0$: $-1\le\cos\tfrac1x\le1$.
- Multiply by $x^2\ (\ge0)$: $-x^2\le x^2\cos\tfrac1x\le x^2$.
- $\displaystyle\lim_{x\to0}(-x^2)=0$ and $\displaystyle\lim_{x\to0}x^2=0$.
- By the Squeeze Theorem, the middle limit is also $0$.
The Squeeze Theorem: if $g\le f\le h$ near $c$ and $\lim g=\lim h=L$, then $\lim f=L$. Use it when a function oscillates or otherwise resists direct methods — bound it above and below by functions with equal limits, verify the inequality near $c$, and $f$ is trapped onto $L$.