Some limits resist every algebra trick because part of the function never settles down. Think of $\sin\left(\frac{1}{x}\right)$, which swings between $-1$ and $1$ infinitely often near 0. The Limit Laws cannot handle it, but a sandwich can.
Key idea
Why it works
Worked example: Taming an oscillation
- The product law fails: $\sin(1/x)$ has no limit at 0.
- But $-1 \le \sin(1/x) \le 1$. Multiply by $x^2 \ge 0$: $-x^2 \le x^2\sin(1/x) \le x^2$.
- Both $-x^2$ and $x^2$ tend to 0.
- By the squeeze theorem the limit is 0.
Answer: $0$
Worked example: Polynomial bounds
- Lower bound: $4(1) - 3 = 1$.
- Upper bound: $1 + 2 - 2 = 1$.
- The bounds agree, so $f$ is squeezed to 1.
Answer: $1$
Worked example: At infinity
- Split: $\frac{3x}{x} + \frac{\cos x}{x} = 3 + \frac{\cos x}{x}$.
- For $x > 0$: $-\frac{1}{x} \le \frac{\cos x}{x} \le \frac{1}{x}$, and both bounds tend to 0.
- So $\frac{\cos x}{x} \to 0$ and the limit is $3 + 0$.
Answer: $3$
Where do the bounds come from?
The most common pattern is bounded times something that goes to 0. Sine and cosine always lie between $-1$ and $1$, so a product like $(x - a)^2\cos(\text{anything})$ is trapped between $-(x - a)^2$ and $(x - a)^2$. More generally, if $\lvert f(x) \rvert \le M$ and $g(x) \to 0$, then $-M\lvert g(x) \rvert \le f(x)g(x) \le M\lvert g(x) \rvert$, so $f(x)g(x) \to 0$. You will use this squeeze again in the next lesson to prove the most important trigonometric limit.
Common mistake