# Squeeze Theorem **The limit of a function at a point is equal to the limits, if they are equal, of an upper and lower bounding function.** ![[SqueezeTheorem.svg|600]] *The function $x^{2}\sin\left(\frac{1}{x}\right)$ being squeezed between $x^{2}$ and $-x^{2}$ at $x=0$; the value of $x^{2}\sin\left(\frac{1}{x}\right)$ at $x=0$ can then be inferred from the value of the other two functions* > [!NOTE] Squeeze Theorem > Let $f$, $g$, and $h$ be functions defined on an interval $I$ containing the point $a$. Suppose that for every $x\neq a$ on the interval, $g(x)\le f(x)\le h(x)$. If the limits of $g(x)$ and $h(x)$ exist and > $\lim_{x\rightarrow a}g(x)=\lim_{x\rightarrow a}h(x)=L$ > then, by the *squeeze theorem*, > $\lim_{x\rightarrow a}f(x)=L$ > [!NOTE]- Example > Suppose the limit > $\lim_{x\rightarrow 0}x^{2}\sin\left(\frac{1}{x}\right)$ > is to be determined. The limit cannot be determined directly because the limit $\lim_{x\rightarrow 0}\sin\left(\frac{1}{x}\right)$ does not exist. > > By the definition of the [[Trigonometric Functions|sine function]], > $-1\le\sin\left(\frac{1}{x}\right)\le 1$ > If the inequality is multiplied through by $x^{2}$, then > $-x^{2}\le x^{2}\sin\left(\frac{1}{x}\right)\le x^{2}$ > As $-x^{2}$ and $x^{2}$ approach zero as $x$ approaches zero, it must also be true that > $\lim_{x\rightarrow 0}x^{2}\sin\left(\frac{1}{x}\right)=0$