AP Calculus BC / Unit 1: Limits and Continuity
1.8 Determining Limits Using the Squeeze Theorem
Notes
If $g(x)\le f(x)\le h(x)$ near $c$ and $\displaystyle \lim g = \lim h = L$, then $\displaystyle \lim f = L$.
- Classic use: $\displaystyle -x^2 \le x^2\sin\frac{1}{x} \le x^2$, so $\displaystyle \lim_{x\to0}x^2\sin\frac{1}{x}=0$.
- It proves $\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$.
Practice
If $2x \le f(x) \le x^2+1$ for all $x$ near 1, what is $\displaystyle \lim_{x\to1}f(x)$?
Both bounds approach 2 at $x=1$.
$\displaystyle \lim_{x\to0}x\cos\left(\frac{1}{x}\right)$ equals:
$-|x|\le x\cos(1/x)\le|x|$, and both sides go to 0.
The squeeze theorem needs the two outer functions to:
If they meet at the same limit, the middle function is trapped there.
Ready for more? Take the Unit 1 test.