The squeeze theorem AB & BC

Suppose that near $x = a$ (excluding $a$ itself)

$$g(x) \le f(x) \le h(x),$$

and that

$$\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L.$$

Then $\lim_{x\to a} f(x) = L$.

There are two conditions and both do real work:

1. The inequality must hold near $a$. Not at one point, and not on a region that excludes one side of $a$ when the limit is two-sided.
2. The two outer limits must exist and be equal. Bounds that close on different values squeeze nothing.

Skipping either check is the characteristic error. Bounds of $-1$ and $1$ around a wildly oscillating function are perfectly true and completely useless, because those bounds do not meet.

Find $\lim_{x\to 0} x^2\sin\left(\frac{1}{x}\right)$. The sine factor oscillates without limit as $x \to 0$, so the product law is unavailable. Build bounds instead.

Start from the one fact that is always true:

$$-1 \le \sin\left(\tfrac{1}{x}\right) \le 1.$$

Multiply through by $x^2$. Because $x^2 \ge 0$, the inequalities keep their direction:

$$-x^2 \le x^2\sin\left(\tfrac{1}{x}\right) \le x^2.$$

Both outer functions tend to 0 as $x\to 0$, so the middle one does too. The answer is 0.

The step everyone skips is the multiplication. Quoting $-1 \le \sin \le 1$ and concluding the limit is 0 leaves the bounds at $-1$ and $1$, which do not meet and prove nothing.

Multiplying an inequality by a negative quantity reverses it, so bounds built by multiplying through are only valid where the multiplier's sign is known.

With $x^2$ there is no issue: it is non-negative everywhere, so a two-sided argument works directly. With a bare $x$ there is: for $x > 0$ multiplying $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ by $x$ gives $-x \le x\sin\left(\frac{1}{x}\right) \le x$, but for $x < 0$ the inequality flips.

The clean fix is to bound the absolute value instead:

$$\left|x\sin\left(\tfrac{1}{x}\right)\right| \le |x|, \quad\text{so}\quad -|x| \le x\sin\left(\tfrac{1}{x}\right) \le |x|,$$

and both $\pm|x|$ tend to 0 from either side. Using a one-sided bound for a two-sided limit is a real error, not a technicality.

Two signals point here. A bounded factor with no limit multiplied by something tending to 0: sines, cosines, and anything oscillating. And a function defined by an inequality rather than a formula, which appears on exams as "suppose $3x - 1 \le f(x) \le x^2 + 1$ near 2".

In the second case, do exactly what the theorem says: take the limit of each bound. Here both give 3 at $x = 2$, so $\lim_{x\to 2} f(x) = 3$, even though $f$ is never actually specified. If instead the bounds gave 3 and 5, the theorem would be silent: the limit could be anything in between, or fail to exist.