Squeeze theorem

What you are looking at

$f(x) = x^2 \sin\dfrac{1}{x}$ oscillates infinitely often near $0$, so no table of values settles its limit. But $\left|\sin\dfrac{1}{x}\right| \le 1$ gives $-x^2 \le x^2\sin\dfrac{1}{x} \le x^2$, and both bounds tend to $0$. The green curves are those bounds, and the working shows $h(x)$, $f(x)$ and $g(x)$ at the current point, always in that order.

The third function is the famous one: for $0 < |x| < \dfrac{\pi}{2}$, $\cos x \le \dfrac{\sin x}{x} \le 1$, and both bounds tend to $1$, so $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$ — the limit behind the derivative of $\sin x$. Zooming does not change the argument; it shows why it works: the closer you look, the less room $f$ has.

Try this

1. 1Press Zoom and watch the window shrink around $0$: $f$ stays between the green curves at every scale.
2. 2Switch to $x \sin\dfrac{1}{x}$: the bounds are now $\pm|x|$, straight lines instead of parabolas, and the squeeze still works.
3. 3Switch to $\dfrac{\sin x}{x}$: it is not defined at $0$, yet the bounds $\cos x$ and $1$ force the limit $1$.
4. 4Hide the formulas and answer the prediction first: does $f$ have a limit at $0$ at all?

Mistakes students make

• Using bounds that do not share a limit. $-1 \le \sin\dfrac{1}{x} \le 1$ is true but proves nothing: the bounds tend to $-1$ and $1$, and $\sin\dfrac{1}{x}$ has no limit at $0$.
• Writing $\lim x^2 \sin\dfrac{1}{x} = \lim x^2 \cdot \lim \sin\dfrac{1}{x}$: the product rule for limits needs both limits to exist, and the second does not.

In the exam

• Squeeze arguments and $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$ are AP Unit 1 material: state both inequalities and both limits before you conclude.
• At A-Level the small-angle approximation $\sin x \approx x$ rests on the same limit.

Where it is taught

• AP Calculus AB/BC · Unit 1 (1.8 Squeeze Theorem)
• IB Maths AA HL · 5.12 (limits)
• Greek Γ΄ Λυκείου · §1.5