Topic 1.8 Determining Limits Using the Squeeze Theorem: apply the squeeze (sandwich) theorem to evaluate limits of functions bounded between two functions with a common limit.

What this topic is asking

The College Board (Topic 1.8) wants you to apply the squeeze theorem (also called the sandwich theorem): if a function is trapped between two other functions that approach the same limit, then it is forced to that limit too. It is the standard way to handle limits of oscillating products like $x^2 \sin\left(\frac{1}{x}\right)$, where no algebraic simplification works.

The theorem

The picture is exactly the name: $f$ is sandwiched between $g$ and $h$. If the bread (the bounds) meets at a single height $L$, the filling ($f$) has nowhere else to go.

How to set it up

The hard part is finding the bounds. The key fact is that bounded oscillating factors stay in a fixed range:

So $x^2 \sin\left(\frac{1}{x}\right)$ is bounded by $-x^2$ and $x^2$, both of which approach $0$.

Why oscillation alone does not kill the limit

A function like $\sin\left(\frac{1}{x}\right)$ has no limit as $x \to 0$ - it oscillates forever between $-1$ and $1$. But multiplying it by $x^2$, which shrinks to $0$, crushes the oscillation down to zero amplitude. The squeeze theorem is the precise tool that proves this intuition.

When to reach for the squeeze theorem

The squeeze theorem is a specialist tool, not a first resort, so it helps to recognize its signature. Reach for it when a limit involves a bounded oscillating factor - typically $\sin$ or $\cos$ of something that blows up, like $\sin\left(\frac{1}{x}\right)$ near $0$ - multiplied by a factor that shrinks to zero. Substitution fails because the oscillating part has no limit, and the algebraic techniques (factoring, conjugates) do not apply because there is nothing to cancel. That combination is the cue. The strategy is then to discard the oscillating factor's exact behavior and only use the fact that it is trapped between $-1$ and $1$, which converts the problem into bounding by two functions you can evaluate. If a limit has no such bounded-times-vanishing structure, the squeeze theorem is usually not the intended method.

What a full justification needs

On an AP free-response question, a complete squeeze argument states the bounding inequality, evaluates the limit of each bound, confirms the two limits are equal, and then names the theorem to conclude. Leaving out the equal-limits check loses credit, because the theorem only works when the bounds meet. A clean template runs: "Since $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$, multiplying by $x^2 \geq 0$ gives $-x^2 \leq f(x) \leq x^2$. As $x \to 0$, both $-x^2 \to 0$ and $x^2 \to 0$, so by the squeeze theorem $\lim_{x \to 0} f(x) = 0$." Reusing this skeleton guarantees you include every element the markers look for.