Statement: If g(x) <= f(x) <= h(x) for all x near a (except possibly at a), and lim(x->a) g(x) = lim(x->a) h(x) = L, then lim(x->a) f(x) = L.

When to use: When the function involves an oscillatory part multiplied by a vanishing part. The classic pattern is $x^n$ * sin$\frac{1}{x}$ or $x^n$ * cos$\frac{1}{x}$ as x -> 0.

Strategy:

1. Identify the bounded oscillatory part: -1 <= sin(anything) <= 1 or -1 <= cos(anything) <= 1
2. Identify the vanishing part that approaches 0
3. Multiply the inequality through by the vanishing part (careful with sign if it can be negative — use |vanishing part|)
4. Both bounds go to 0, so by Squeeze Theorem, the function also goes to 0

Examples:

• lim(x->0) x * sin$\frac{1}{x}$ = 0 (bounded between -|x| and |x|)
• lim(x->0) $x^2$ * cos(1/$x^3$) = 0 (bounded between -$x^2$ and $x^2$)
• lim(x->infinity) sin$\frac{x}{x}$ = 0 (bounded between -1/x and 1/x)

JEE Application: Often appears in assertion-reasoning format or as a step within a larger problem. Also used to prove that lim(x->0) sin$\frac{x}{x}$ = 1 geometrically.